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**NCERT Exemplar Class 11 Maths Solutions Chapter 8** - Squares and roots solving have always been a major issue for a lot of students when solving algebra. For those who cannot use calculators in school, French mathematician Blaise Pascal removed the Binomial Theorem to make the process easier and faster. In NCERT exemplar Class 11 Maths solutions chapter 8, we discover a term known as the Pascal Triangle, which deals with how array arrangements of coefficients of expressions are for the derivation of the root answer. Integers and indices play an important role in this NCERT exemplar Class 11 Maths solutions chapter 8 where they are kept in formulas to find out answers. The positioning of the terms in the binomial theorem has a value and its importance, which is explained well in the General and Middle terms section. NCERT exemplar Class 11 Maths solutions chapter 8 pdf download can be availed using ‘webpage download as PDF online tools.**Also, check - ** NCERT Class 11 Maths Solutions for other chapters

**JEE Main Scholarship Test Kit (Class 11): Narayana | Physics Wallah | **

**Suggested: ****JEE Main: high scoring chapters | ****Past 10 year's papers**

Question:1

Find the term independent of x, where x≠0, in the expansion of

Answer:

............... (Given)

……. (from standard formula of Tr+1)

…… (i)

Now, for x,

30 – 3r = 0

→ r = 10

Substituting the value of r in eq (i),

Question:2

if the term free from x is the expansion of is 405, then find the value of k.

Answer:

................... (Given)

……. (from standard formula of Tr+1)

…….. (i)

Now, for x,

(10 – 5r)/2 = 0

→ r = 2

Substituting the value of r in eq (i),

= 405

Thus k^{2} = 9

Thus, k = ±3

Question:3

Find the coefficient of x in the expansion of

Answer:

(1-3x+7x^{2}) (1-x)^{16 } ….. (given)

= (1-3x+7x^{2}) (^{106}C_{0 }- ^{16}C_{1}x^{1} + ^{16}C_{2}x^{2} + …… + ^{16}C_{16}x^{16})

= (1-3x +7x^{2}) (1-16x+120x^{2}+…..)

Thus, coefficient of x = -19

Question:4

Find the term independent of x in the expansion of

Answer:

Given: (3x – 2/x^{2})15

Thus,

Now, for independent term x,

15 – 3r = 0

Thus, r = 5

Thus, the term x is

T_{5+1} = ^{15}C_{5}3^{15-5}(-2)^{5}

= -15 x 14 x 13 x 12 x 11 x 10!/5x4x3x2x1x10! .3^{10}.2^{5}

= -3003 x 3^{10}x 2^{5}

Question:5

Find the middle term (terms) in the expansion of

i) ii)

Answer:

(i) Given: (x/a – a/x)^{10}

Where, n = 10(even).

The middle term = (10/2 + 1)th term, viz., the sixth term

Thus, T_{6} = T_{5+1 }^{10}C_{5} (x/a)^{10-5} (-a/x)^{5}

= - ^{10}C_{5} (x/a)^{5}(a/x)^{5}

= - 10 x 9 x 8 x 7 x 6 x 5!/ 5! x 4 x 3 x 2 x 1 (x/a)^{5}(x/a)^{-5}

= -252

(ii) Given: (3x – x^{3}/6)^{9}

Where, n = 9 (odd)

Now, here are two middle terms-

(9+1/2)^{th} ,viz., the tenth term

& (9+1/2 + 1)^{th}, viz., the sixth term

Thus, T_{5} = T_{4+1}

= ^{9}C_{4} (3x)^{9-4} (-x^{3}/6)^{4}

= 9x8x7x6x5!/4x3x2x1x5! . 3^{5}x^{5}x^{12}6^{-4}

= 189/8 x^{17}

Now,

T_{6} = T_{5+1}

= ^{9}C_{5}(3x)^{9-5} (-x^{3}/6)^{5}

= - 9 x 8 x 7 x 6 x 5!/ 5! x 4 x 3 x 2 x 1 . 3^{4}.x^{4}.x^{15}.6^{-4}

= - 21/16 x^{19}

Question:6

Find the coefficient of x^{15} in the expansion of (x – x^{2})^{10}

Answer:

Given: (x – x^{2})^{10}

Thus, T_{r+1} = ^{10}C_{r} x^{10-r}(-x^{2})^{r}

= (-1)^{r 10}C_{r}x^{10-r}x^{2r}

= (-1)^{r 10}C_{r}x^{10+r}

Now, for coefficient of x^{15}

We have,

10 + r = 15

Thus, r = 5

Thus, T_{5+1} = (-1)^{5}.^{10}C_{5}x^{15}

Therefore,

Coefficient of x^{15} = -10x9x8x7x6x5!/5x4x3x2x1x5!

= -252

Question:7

Find the coefficient of x^{1/17} in the expansion of (x^{4} – 1/x^{3})^{15}

Answer:

Given: (x^{4} – 1/x^{3})^{15}

Thus,

T_{r+1 }= ^{15}C_{r}(x^{4})^{15-r}(-1/x^{3})^{r}

= ^{15}C_{r} x^{60-4r}(-1)^{r} x^{-3r}

= ^{15}C_{r} x^{60-7r}(-1)^{r}

Now, for coefficient x^{-17},

60 – 7r = -17

7r = 77

r = 11

Thus, T_{11+1} = ^{15}C_{11} x^{60-77}(-1)^{11}

Therefore,

Coefficient of x^{-17} = -15 x 14 x 13 x 12 x 11!/11! x 4 x 3 x 2 x 1

= -1365

Question:8

Given: (y^{1/2} + x^{1/3})^{n}

From the end, binomial coefficient of 3^{rd} term = 45

Thus, ^{n}C_{n-2} = 45 , ^{n}C_{2 }= 45

i.e., n(n-1)(n-2)!/2!(n-2)! = 45

Now, n(n-1) = 90,

n^{2} – n – 90 = 0,

(n – 10)(n + 9) = 0

n = 10 …. (since n is not equal to -9)

Now, the 6^{th} term,

=^{10}C_{5} (y^{1/2})^{10-5 }(x^{1/3})^{5}

= 252y^{5/2}.x^{5/3}

Question:9

Answer:

Given: (1+x)^{18}

Now, T_{(2r+3)+1 }is the (2r + 4)^{th} term

Thus, T_{(2r+3)+1} = ^{18}C_{2r+3} (x)^{2r+3}

Now, T_{(r-3)+1} is the (r-2)^{th} term

Thus, T_{(r-3)+1} = ^{18}C_{r-3} x^{r-3}

Now, it is given that the terms are equal in expansion,

Thus,^{18}C_{2r+3 }= ^{18}C_{r-3}

i.e., 2r + 3 + r – 3 = 18 …… (^{n}C_{x} = ^{n}C_{y} gives x + y = n)

Thus, 3r = 18

Therefore, r = 6

Question:10

Given: (1+x)^{2n}

Now, ^{2n}C_{1}, ^{2n}C_{2}, ^{2n}C_{3} are the coefficients of the first second and the third terms respectively.

It is also given that the coefficients are in A.P.

Thus, 2. ^{2n}C_{2 }=^{2n}C_{1} +^{ 2n}C_{3}

Thus, 2[2n(2n-1)(2n-2)!/2x1x(2n-2)!] = 2n + 1n(2n-1)(2n-2)(2n-3)!/3!(2n-3)!

N(2n-1) = n = n(2n-1)(n-1)/3

3(2n-2) = 3+ (2n^{2} – 3n + 1)

6n – 3 = 2n^{2} – 3n + 4

Thus, 2n^{2} – 9n + 7 = 0

Question:11

Find the coefficient of x^{4} in the expansion of (1 + x + x^{2} + x^{3})^{11.}

Answer:

Given: (1+x+x^{2}+x^{3})^{11}

That is equal to [(1+x) + x^{2(}1 + x)]^{11}

= [(1+x) (1+x^{2})]^{11}

= (1+x)^{11}.(1+x^{2})^{11}

= (^{11}C_{0} + ^{11}C_{1}x + ^{11}C_{2}x^{2} + ^{11}C_{3}x^{3} +^{ 11}C_{4}x^{4} + ….) (^{11}C_{0} + ^{11}C_{1}x^{2} + ^{11}C_{2}x4 + ….)

Thus, the coefficient of x^{4} =^{11}C_{0} x ^{11}C4 + ^{11}C1 x ^{11}C2+^{11}C2x ^{11}C_{0}

= 330 + 605 + 55

= 990

Question:12

Answer:

Given: (p/2 + 2)^{8}

Now, here index is n = 8,

Since there is only one middle term, viz., (8/2 + 1)th i.e., the fifth term

T_{5} = T_{4+1}

= ^{8}C_{4}(p/2)^{8-4}. 2^{4}

Now,

1120 = ^{8}C_{4}p^{4}

1120 = 8x7x6x5x4!/4!x4x3x2x1 . p^{4}

1120 = 7x2x5xp^{4}

Thus, p^{4 }= 1120/70

p^{4}= 16

p^{2} = 4

Thus, p = ±2

Question:13

Show that the middle term in the expansion ofis

Answer:

Given: (x – 1/x)^{2n}

Now, here the index = 2n(even)

There’s only one middle term, i.e., (2n/2 + 1)^{th }term, viz., (n + 1)^{th }term

T_{n+1} = ^{2n}C_{n}(x)^{2n-n}(-1/x)^{n}

= ^{2n}C_{n}(-1)^{n}

= (-1)^{n}(2n!)/n!.n!

= (-1)^{n} 1.2.3.4.5. …. (2n-1).(2n)/n!.n!

= (-1)^{n} [1.3.5…. (2n-1)].2^{n}[1.2.3…..n]/(1.2.3…n).n!

= (-2)^{n}[1.3.5…(2n-1)].2^{n}/n!

Question:14

Find n in the binomial if the ratio of 7th term from the beginning to the 7th term from the end is

Answer:

Given

Now, from the beginning,

Seventh term is

...................(i)

& from the end,

The seventh term is the same ,i.e.,

.............(ii)

Now, it is given that

Thus

Thus

Thus

Thus n=9

Question:15

Answer:

(i) (x+a)^{n}= ^{n}C_{0}x^{n} + ^{n}C_{1}x^{n-1} a^{1} + ^{n}C_{2}x^{n-2} a^{2} + ^{n}C_{3}x^{n-3} a^{3} + ….. + ^{n}C_{n} a^{n}

Now, sum of odd terms,

O = ^{n}C_{0}x^{n}+ ^{n}C_{2}x^{n-2} a^{2} + …….

& sum of even terms,

E = ^{n}C_{1}x^{n-1} a^{1} + ^{n}C_{3}x^{n-3} a^{3} + …..

Now, (x+a)^{n} = O+E & (x-a)^{n} = O – E

Thus, (O + E)(O – E) = (x + a)^{n}(x – a)^{n}

Thus, O^{2} – E^{2} = (x^{2} – a^{2})^{n}

(ii)Here,

4OE = (O + E)^{2} – (O – E)^{2}

= [(x+a)^{n}]^{2} – [(x-a)^{n}]^{2}

= (x+a)^{2n} – (x-a)^{2n}

Question:16

If x^{n } occurs expantion of then prove that its coefficient is

Answer:

Given: (x^{2} + 1/x)^{2n}

Thus, T_{r+1 }=^{2n}C_{r}(x^{2})^{2n-r}(1/x)^{r}

= ^{2n}C_{r}x^{4n-3r}

Now, if xn will occur in the expansion, then,

Let us consider 4n – 3r = p

Thus,

3r = 4n – p

r = 4n – p/3

Thus, coefficient of xp = ^{2n}C_{r}

= (2n)!/r!(2n-r)!

=

Question:17

Find the term independent of x in the expansion of(1+x+2x^{3})(3x^{2}/2 – 1/3x)^{9}

Answer:

Given: (1+x+2x^{3})( 3x^{2}/2 – 1/3x)^{9}

Now, let us consider (3x^{2}/2 – 1/3x)^{9}

T_{r+1 }= ^{9}C_{r }(3/2 x^{2})^{9-r} (-1/3x)^{r}

= ^{9}C_{r} (3/2)^{9-r} (-1/3)^{r }x^{18-3r}

Thus, in the expansion of (1+x+2x^{3})(3/2x^{2} – 1/3x)^{9} the general term is:^{9}C_{r }(3/2)^{9-r} (-1/3)^{r} x^{18-3r }+ ^{9}C_{r} (3/2)^{9-r }(-1/3)^{r} x^{19-3r} + 2.^{9}C_{r} (3/2)^{9-r }(-1/3)^{r} x^{21-3r}

Now, for the independent term x,

Let us put 18 – 3r = 0,

19 – 3r = 0

& 21 – 3r = 0,

We get,

R = 6 & r = 7

Now, since the second term is not independent of x, it will be:^{9}C_{6} (3/2)^{9-6} (-1/3)^{6} + 2.^{9}C_{r} (3/2)^{9-7} (-1/3)^{7}

= 9x8x7x6!/6!x3x2 . 1/2^{3}.3^{3} – 2. 9x8x7!/7!x2x1. 3^{2}/2^{2}.1/3^{7}

= 84/8.1/3^{3} – 36/4.2/3^{5}

= 17/54

Question:18

The total number of terms in the expansion of (x + a)^{100 }+ (x – a)^{100 }after simplification is

(A) 50 (B) 202 (C) 51 (D) none of these

Answer:

(c) 51.

Now, (x+a)^{100 }+ (x-a)^{100} = (^{100}C_{0}x^{100} + ^{100}C_{2}x^{99}a + ^{100}C_{2}x^{98}a^{2} + ….) + (^{100}C_{0}x^{100 }- ^{100}C_{2}x^{99}a + ^{100}C_{2}x^{98}a^{2})

= 2(^{100}C_{0}x^{100} + ^{100}C_{2}x^{98}a^{2 }+…..+^{100}C_{100}a^{100})

Therefore, there are 51 terms.

Question:19

Given the integers r > 1, n > 2, and coefficients of (3r)^{th} and (r + 2)^{nd} terms in the binomial expansion of (1 + x)^{2n} are equal, then

(A) n = 2r (B) n = 3r (C) n = 2r + 1 (D) none of these

Answer:

(a) n =2r

Given: (1+x)^{2n}

Thus, T_{3r} = T_{(3r-1)+1}

= ^{2n}C_{3r-1}x^{3r-1}

& T_{r+2 }= T_{(r+1)+1}

= ^{2n}C_{r+1}x^{r+1}

Now, ^{2n}C_{3r-1}x^{3r-1} = ^{2n}C_{r+1}x^{r+1} ……… (given)

Thus, 3r – 1 + r + 1 = 2n

Thus, n = 2r

Question:20

The two successive terms in the expansion of (1 + x)^{24} whose coefficients are in the ratio 1: 4 are

(A) 3^{rd} and 4^{th} (B) 4^{th} and 5^{th} (C) 5^{th} and 6^{th} (D) 6^{th} and 7^{th}

Answer:

(c) 5^{th} & 6^{th}

Let us consider (r+1)^{th}& (r+2)^{th} as the 2 successive terms in the expansion of (1+x)^{24}

Now, T_{r+1} = ^{24}C_{r}x^{r}& T_{r+2} = ^{24}C_{r+1}x^{r+1}

Now, ^{24}C_{r}/^{24}C_{r+1} = ¼ ……. (given)

Thus,

Thus, (r+1)r! (23 – r)! / r!(24 – r)(23 – r)! = ¼

r + 1/24 – r = ¼

4r + 4 = 24 – r

t = 4

Thus, T_{4+1} = T_{5}

& T_{4+2} = T_{6}

Thus, opt (c).

Question:21

The coefficients of x^{n} in the expansion of (1 + x)^{2n} and (1 + x)^{2n ~1} are in the ratio

(a) 1 : 2

(b) 1 : 3

(c) 3 : 1

(d) 2:1

Answer:

The answer is the option (d) 2:1^{2n}C_{n} is the coefficient of x^{n} in the expansion of (1+x)^{2n}

&^{2n-1}C_{n} is the coefficient of x^{n }in the expansion of (1+x)^{2n-1}

= (2n)!n!(n-1)!/(2n-1)!n!n!

= 2n(2n-1)!n!(n-1)!/n!n(n-1)!(2n-1)!

= 2/1

= 2:1

Question:22

If the coefficients of 2^{nd}, 3^{rd} and the 4^{th} terms in the expansion of (1 + x)^{n} are in A.P., then the value of n is

(a) 2

(b) 7

(c) 11

(d) 14

Answer:

The answer is the option (b) 7

(1+x)^{n} = ^{n}C_{0 }+ ^{n}C_{1}x + ^{n}C_{2}x^{2} +^{ n}C_{3}x^{3} + ….. + ^{n}C_{n}x^{n}

Now, ^{n}C_{1}, ^{n}C_{2}, ^{n}C_{3} are the coefficients of the second, third and fourth terms respectively.

2 ^{n}C_{2} =^{n}C_{1} + ^{n}C_{3} …… since they are in A.P.

2[n!/(n-2)!2!] = n + n!/ 3!(n-3)!

2[n(n-1)/2!] = n + n(n-1)(n-2)/3!

(n-1) = 1 + (n-1)(n-2)/6

6n – 6 = 6 + n^{2} – 3n + 2

n^{2} – 9n + 14 = 0

(n-7)(n-2) = 0

Thus, n = 2 or n = 7

Now, n = 2 is not possible,

Thus, n = 7

Question:23

If A and B are coefficients of x^{n } in the expansions of (1 + x)^{2n} and (1 + x)^{2n–1} respectively, then A/B equals to

(a) 1

(b) 2

(c)

(d)

Answer:

The answer is the option (b) 2^{2n}C_{n i}s the coefficient of xn in the expansion of (1+x)^{2n}

Thus, A = ^{2n}C_{n}

&^{2n-1}C_{n} is the coefficient of x^{n} in the expansion of (1+x)^{2n-1}

Thus, B =^{ 2n-1}C_{n}

Thus, A/B = ^{2n}C_{n}/^{2n-1}C_{n}

= 2/1

=2:1

Question:24

If the middle term of is equal to then the value of x is

(a) (b) (c) (d)

Answer:

The answer is the option (c) nπ + (-1)^{n}π/6

Now, n = 10 (even)

Thus, there is only one middle term i.e., the 6^{th} term

Thus, T_{6} = T_{5+1}

= ^{10}C_{5}(1/x)^{10-5} (x sin x)^{5}

63/8 = ^{10}C_{5 }sin^{5}x …… (given)

63/8 = 252 x sin^{5}x

Thus, sin^{5}x = 1/32

Sin x = ½

Sin x = π/6

Thus, x = nπ + (-1)^{n}π/6

Question:25

The largest coefficient in the expansion of (1 + x)^{30} is _________________ .

Answer:

^{30}C_{30/2} = ^{30}C_{15} is the largest coefficient in the expansion of (1+x)^{30}

Question:27

In the expansion of (x^{2} – 1/x^{2})^{16}, the value of constant term is_________________.

Answer:

Here, T_{r+1 }= ^{16}C_{r}(x^{2})^{16-r}(-1/x^{2})^{r}

= ^{16}C_{r}x^{32-4r}(-1)^{r}

Now, for constant term,

32 – 4r = 0

Thus, r = 8

Thus, T_{8+1} = ^{16}C_{8}

Question:28

Answer:

Given:

Now, referring to que 14.

Thus

Thus

6^{(}^{n-12)/3} = 6^{0}

(n-12)/3 = 0

Thus, n = 12

Question:29

The coefficient of in the expansion of is ______________

Answer:

Given: (1/a – 2b/3)^{10}

Thus, T_{r+1 }= ^{10}C_{r}(1/a)^{10-r} (-2b/3)^{r}

Now, for the coefficient of a^{-6}b^{4},

10 – r = 6

Thus, r = 4

Therefore the coefficient of a^{-6}b^{4} = ^{10}C_{4}(-2/3)^{4}

= 10.9.8.7.6!/6!.4.3.2.1 . 2^{4}/3^{4}

= 1120/27

Question:30

Middle term in the expansion of (a^{3} + ba)^{28}is _________

Answer:

Given: (a^{3} + ba)^{28}

Here, n = 28 (even)

Thus, there is only one middle term (28/2 + 1)th or 15^{th} term.

Thus, the middle term, T_{15} = T_{14+1}

= ^{28}C_{14}(a^{3})^{28-14}(ba)^{14}

= ^{28}C_{14}a^{42}b^{14}a^{14}

= ^{28}C_{14}a^{56}b^{14}

Question:31

The ratio of the coefficients of x^{p} and x^{q} in the expansion of (1 +x)^{p+q} is _______

Answer:

Given: (1+x)^{p+q}

Thus, coefficient of x^{p} = ^{p+q}C_{p}

& Coefficient of x^{q} = ^{p+q}C_{q}

Therefore,^{p+q}C_{p}/ ^{p+q}C_{q} = ^{p+q}C_{p} /^{p+q}C_{p}

= 1:1

Question:32

The position of the term independent of x in the expansion of is _____________.

Answer:

Given:

Thus,

Now, for constant term,

10 – 5r = 0

Thus, r = 2

Therefore, the third term is independent of x.

Question:33

If 25^{15} is divided by 13, the reminder is _________ .

Answer:

Now, 25^{15 }= (26 – 1)^{15}

= ^{15}C_{0}26^{15 }– ^{15}C_{1}26^{14} + …. + ^{15}C_{14}26 – ^{15}C_{15}

= (^{15}C_{0}26^{15} – ^{15}C_{1}26^{14} + …. + ^{15}C_{14}26 – 13) + 12

Therefore, it is clear that, when we divide 25^{15 }by 13, we get 12 as the remainder.

Question:34

Answer:

The given statement is False.

Now, from what’s given, we can say that,

= ^{20}C_{0} + ^{20}C_{1} + ^{20}C_{2} +^{ 20}C_{3} + …… + ^{20}C_{10}

=^{ 20}C_{0 }+ ^{20}C_{1} + ….. + ^{20}C_{10} +^{ 20}C_{11} + …… + ^{20}C_{20} – (^{20}C_{11} + ….. + ^{20}C_{20})

= 2^{20 }– (^{20}C_{11} + …… + ^{20}C_{20})

Question:35

The expression 7^{9} + 9^{7} is divisible by 64.

Answer:

The given statement is True.

Given: 7^{9} + 9^{7} is divisible by 64.

Now,

7^{9} + 9^{7} = (1+8)^{7} – (1-8)^{9}

= [^{7}C_{0 }+^{ 7}C_{1}.8 + ^{7}C_{2}.(8)^{2} + ^{7}C_{3}.(8)^{3} + ….. + ^{7}C_{7}.(8)^{7}] – [^{9}C_{0} + ^{9}C_{1}.8 + ^{9}C_{2}.(8)^{2} - ^{9}C_{3}.(8)^{3} + ….. ^{9}C_{9}.(8)^{0}]

= (7x8 + 9x8) + (21 x 8^{2} – 36 x 8^{2}) + …..

= 128 + 64(21 – 36)

= 64[2 + (21 – 36) + …], viz. divisible by 64.

Question:36

The number of terms in the expansion of [(2x + y^{3})^{4}]^{7} is 8.

Answer:

The given statement is False.

Given: [(2x + y^{3})^{4}]^{7} = (2x + y^{3})^{28}

Thus, no. of terms = 28 + 1

= 29

Question:37

Answer:

The given statement is False.

Given: (1 + x)^{2n-1}

Now, no. of terms = 2n – 1 + 1

= 2n (even)

Thus, the middle term = 2n/2^{th }terms & (2n/2 + 1)^{th} terms

= n^{th }terms & (n+1)^{th} terms

Now, the coefficient of-

n^{th} term =^{ 2n – 1}C_{n-1}

& (n + 1)^{th} term = ^{2n-1}C_{n}

Now, sum of the coefficients = ^{2n – 1}C_{n-1 }+ ^{2n-1}C_{n}

= ^{2n-1}C_{n-1} + ^{2n-1}C_{n}

=^{ 2n}C_{n}

Question:38

The last two digits of the numbers 3^{400} are 01.

Answer:

The given statement is true.

Given: 3^{400 }= (9)^{200}

= (10 – 1)^{200}

Thus, (10 – 1)^{200} = ^{200}C_{0}(10)^{200} – ^{200}C_{1}(10)^{199 }+ ….. – ^{200}C_{199}(10)^{1} +^{ 200}C_{200}(1)^{200}

= 10^{200} – 200 x 10^{199} + …. – 10 x 200 + 1

Therefore, the last two digits will be 01.

Question:39

If the expansion of (x-1/x^{2})^{2n} contains a term independent of x, then n is a multiple of 2.

Answer:

The given statement is False.

Given:

Now,

T_{r+1} = ^{2n}C_{r}(x)^{2n-r} (-1/x^{2})^{r}

= ^{2n}C_{r(}x)^{2n-r-2r}(-1)^{r}

= ^{2n}C_{r}(x)^{2n-3r} (-1)^{r}(-1)^{r}

Now, for the independent term of x,

2n – 3r = 0

Thus, r = 2n/3, viz. not an integer

Thus, this expression is not possible to be true.

Question:40

Number of terms in the expansion of (a + b)^{n} where n ∈ N is one less than the power n.

Answer:

The given statement is False.

In the given expression (a+b)^{n}, the no. of terms is just 1 more than n, i.e., n + 1.

- 8.1 Introduction
- 8.2 Binomial Theorem for Positive Integral Indices
- 8.2.1 Binomial theorem for any positive integer n
- 8.2.2 Some special cases
- 8.3 General and Middle Terms

**NCERT Exemplar Class 11 Maths Solutions Chapter-Wise**

Through NCERT Exemplar Class 11 Maths solutions chapter 8 will help students easily understand how to calculate roots and squares, with the help of the theorem. The slant array presentation will help them understand the pascal triangle with much ease and interest.

In NCERT exemplar solutions for class 11 Maths chapter 8, you will learn a very important statement about the theorem, i.e. that any positive integer can be termed as ‘n’, there will be a sum of any two numbers, of which n will be the power and how the sum of n+1 will derive. This is actually how the binomial theorem came into existence and is solved by people across. These solutions will ease the difficulty in finding root for students without the help of a calculator.

· Class 11 Maths NCERT Exemplar solutions chapter 8 has detailed that Squares and roots, binomial theorems, special cases and special terms are important topics which students should pay extra attention to.

· They will learn squares and roots with ease and score well in class. Major and complex problems can be easily solved with the help of this theorem. For students who want to pursue higher studies in this field should be very thorough while studying this chapter.

**Check Chapter-Wise NCERT Solutions of Book**

Chapter-1 | |

Chapter-2 | |

Chapter-3 | |

Chapter-4 | |

Chapter-5 | |

Chapter-6 | |

Chapter-7 | |

Chapter-8 | Binomial Theorem |

Chapter-9 | |

Chapter-10 | |

Chapter-11 | |

Chapter-12 | |

Chapter-13 | |

Chapter-14 | |

Chapter-15 | |

Chapter-16 |

**Read more NCERT Solution subject wise -**

- NCERT Solutions for Class 11 Maths
- NCERT Solutions for Class 11 Physics
- NCERT Solutions for Class 11 Chemistry
- NCERT Solutions for Class 11 Biology

**Also, read NCERT Notes subject wise -**

- NCERT Notes for Class 11 Maths
- NCERT Notes for Class 11 Physics
- NCERT Notes for Class 11 Chemistry
- NCERT Notes for Class 11 Biology

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1. Are these solutions helpful in entrance exams?

Yes, these solutions help prepare the topics from base-up during entrance exam preparation.

2. How to use these solutions?

One can use these NCERT Exemplar Class 11 Maths chapter 8 solutions as a reference point while solving questions on one’s own.

3. Are the solutions for free?

Yes, these solutions are available for free as one can download them directly from the link provided.

4. Are these solutions as per the NCERT pattern?

Yes, these NCERT Exemplar Class 11 Maths solutions chapter 8 are solved as per the NCERT pattern in the most detailed way as per the marking scheme.

Sep 06, 2024

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