NCERT Grade 11-Binomial Theorem-Answers

NCERT Solutions for Class 11 Maths

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1.    Using Binomial Theorem,

(1 – 2x)5 = 5C0 + 5C1 (-2x) + 5C2 (-2x) + 5C3 (-2x) + 5C4 (-2x) + 5C(-2x)

= 1 + 5(-2x) + 10(-2x)2 + 10(-2x)3 + 5(-2x)4 + (-2x)5

= 1 + 5(-2x) + 10(4x2) + 10(-8x3) + 5(16x4) + (-32x5)

= 1 – 10x + 40x2 – 80x3 + 80x4 – 32x5.

2.    Using Binomial Theorem,

3.    Using Binomial Theorem,

(2x – 3)6 = 6C0(2x)6 + 6C1 (2x)5 (-3) + 6C2 (2x)4 (-3)2 + 6C3 (2x)3 (-3)3 + 6C4 (2x)2 (-3)4 + 6C(2x) (-3)5 + 6C(-3)6

= 64x6 + 6(32x5) (-3) + 15 (16x4) (9) + 20(8x3) (-27) + 15 (4x2) (81) + 6(2x) -(243) + 7

= 64x6 – 576x5 + 2160x4 – 4320x3 + 4860x2 – 2916x + 729

4.    Using Binomial Theorem,

5.    Using Binomial Theorem,

6.   First we have to express 96 as the sum or difference of two numbers whose powers are easier to calculate and then use Binomial Theorem

We can write 96 = 100 – 4

Therefore (96)3 = (100 – 4)3

Using Binomial Theorem,

= (100 – 4)3 = 3C0(100)3 + 3C1(100)2 (-4) + 3C2(100) (-4)2 + 3C3(-4)3

= (100)3 + 3.1000(-4) + 3.100.16 + (-64)

= 1000000 – 120000 + 4800 – 64

= 1004800 – 120064 = 884736

7.    First we have to express 102 as the sum or difference of two numbers whose powers are easier to calculate and then use Binomial Theorem

We can write 102 = 100 + 2

Therefore (102) = (100 + 2)

Using Binomial Theorem,

(100 + 2)5 = 5C0(100)5 + 5C1(100)4 (2) + 5C2(100)2 (2)2 + 5C3(100)2(2)3 + 5C4(100)(2)4

= (100)5 +5 (100)4(2) + 10 (100)3 (2)2 + 10(100)2 (2)3 + 5.100 (2)4(2)5

= 10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32 = 11040808032.

8.   First we have to express 101 as the sum or difference of two numbers whose powers are easier to calculate and then use Binomial Theorem

We can write 101 = 100 + 1

Therefore (101)4 = (100 + 1)4

Using Binomial Theorem,

= (100 + 1)4 = 4C0(100)4 + 4C1(100)3 (1) + 4C2(100)2 (1)2 + 4C3(100)1(1)3 + 4C4(1)4

= (100)4 + 4 (100)3 + 6 (100)2 + 4(100) + 1

= 100000000 + 4000000 + 60000 + 400 + 1 = 104060401.

9.   First we have to express 99 as the sum or difference of two numbers whose powers are easier to calculate and then use Binomial Theorem

We can write 99 = 100 – 1

Therefore (99)5 = (100 – 1)5

Using Binomial Theorem,

(100 – 1)5 = 5C0(100)5 + 5C1(100)4 (-1) + 5C2(100)3 (-1)2 + 5C3(100)2(-1)3 + 5C4(100)(-1)+ 5C5(-1)5

= (100)5 + 5 (100)4 (-1) + 10 (100)3 (1) + 10(100)2 (-1) + 5(100) (1) + (-1)

= 10000000000 – 500000000 + 10000000 – 100000 + 500 – 1 = 9509900499.

10.   We have 1.1 = 1 + 0.1

∴ (1.1)10000 = (1 + 0.1)10000

Using Binomial Theorem,

(1 + 0.1)10000 = 1 +10000 C1 (.1) +10000 C2(0.1)2 +10000 C3(0.1)3 + …..

= 1 + 10000 (0.1) + other positive numbers

= 1 + 1000 + other positive numbers

which is greater than 1000

∴ (1.1)10000 > 1000

11.   Given: (a + b)4 – (a – b)4

Using Binomial Theorem,

(a + b)4 – (a – b)4 = [4C0a4 + 4C1a3b + 4C2a2b2 + 4C3ab3 + 4C4b4]

-[4C0a4 + 4C1a3 (-b) + 4C2a2 (-b)2 + 4C3a(-b)3 + 4C4(-b)4]

= a4 + 4a3b + 6a2b2 + 4ab3 + b4 – [a4 – 4a3b + 6a2b2 – 4ab3 + b4]

= a4 + 4a3b + 6a2b2 + 4ab3 + b4 – a4 + 4a3b – 6a2b2 + 4ab3 – b4

= 8a3b + 8ab3 = 8ab(a2 + b2)

12.   Given: (x + 1)6 + (x – 1)6

Using Binomial Theorem,

13.   We know that b is divisible by a (or a divides b) ⇒ b = ak, k is an integer

Here we have to show that 64 divides 9n+1 – 8n – 9

⇒ 9n+1 – 8n – 9 = 64k, k is an integer

We have 9n+1 = (1 + 8)n+1

Using Binomial Theorem,

14.   

15.   The general term in the binomial expansion of (x + a)is given by Tr+1 = Cn-r ar

∴ General term in the expansion of (x + 3)8 is Tr+1 = 8Crx8-r (3)r      ……(i)

Comparing the indices of x in x5 and in Tr+1

we get 8 – r = 5

=> r = 3

Putting r = 3 in eq. (i), T4 = 8C3x8-3 (3)= 8C3x5 (3)r

Therefore, coefficient of x5 on the expansion (x + 3)8 is 8C3(3)3 = 56 × 27 = 1512.

16.   The general term in the binomial expansion of (a + b)n is given by Tr+1 = nCran-r br

General term in the expansion of = (a – 2b)12 = [a + (-2b)12] is

Tr+1 = 12Cra12-r (-2b)r = 12Cr(-2)r a12-rbr        …(i)

Comparing the indices of a and b in a5b7 and in Tr+1 we get r = 7

Putting r= 7 in eq. (i), T8 = 12C7a5 (-2b)7 = 12C7(-2)7 a5b7

Therefore, coefficient of a5b7 on the expansion (a – 2b)12 is 12C7(-2)7 = -101376.

17.   The general term in the binomial expansion of (x + a)n is given by Tr+1 = nCrxn-r ar

∴ General term in the expansion of (x2 – y)6 = [x2 + (y)]6 is

Tr+1 = 6Cr(x2)6-r (-y)r

⇒ Tr+1 = (-1)r 6Crx12-2r yr

18.   The general term in the binomial expansion of (x + a)n is given by Tr+1 = nCrxn-r ar

∴ General term in the expansion of (x2 – yx)12 is  Tr+1 = 12Cr(x2)12-r (-yx)r

⇒ Tr+1 = (-1)r 12Crx24-2r yrxr = (-1)r 12Crx24-r yr

19.   The general term in the binomial expansion of (x + a)n is given by Tr+1 = nCrxn-r ar

∴ General term in the expansion of (x – 2y)12 is  Tr+1 = 12Cr(x)12-r (-2y)r

⇒ Tr+1 = (-1)r 12Cr2rx12-2r yr

Putting r = 3, T4 = (-1)3 12C3.23x12-3 y3

⇒ T4 = – 12C3.8x9y= -220 × 8x9y= -1760x9y3

20.  The general term in the binomial expansion of (x + a)n is given by Tr+1 = nCrxn-r ar

21.   We have if n is odd, then n+1 is even, so there will be two middle terms in the expansion of (a + b)n, namely th term

Here n = 7, which is an odd number.

Therefore, the middle terms are  i.e., 4th and 5th terms.

The general term in the binomial expansion of (a + b)n is Tr+1 = nCran-r br

22.   We have if n is even, then n + 1 is odd, so there will be a single middle term in the expansion of (a + b)n, namely th term.

Here n = 10, which is an even number.

Therefore, the middle term is  term i.e., 6th term.

The general term in the binomial expansion of (a + b)n is Tr+1 = nCran-r br

23.   We know that,

(1 + x)n = nC0 + nC1x + nC2x2 + nC3x3+…………+ nCnxn and general term of this expansion is Tr+1 = nCrxn

∴ General term in the expansion of (1 + a)m+n will be Tr+1 = m+nCrar

Now, coefficient of am in the expansion of (1 + a)m+n = m+nCm and

Coefficient of an in the expansion of (1 + a)m+n = m+nCn

But we have nCrnCn-r

∴ Coefficient of am = m+nCm = m+nC(m+n) = m+nCn = coefficient of an

Hence proved.

24.   We know that,

(1 + x)n = nC0 + nC1x + nC2x2 + nC3x3+…………+ nCnxn and general term of this expansion is Tr+1 = nCrxn

Therefore the coefficient of the (r – 1)th, rth and (r + 1)th terms are nCr-2, nCr-1 and nCr respectively.

Multiplying equation (ii) by 2 and subtracting it from equation (i), we get n – 7 = 0 ⇒ n = 7

Now substituting n = 7 in equation (i) we get 24 – 8r = 0 ⇒ r = 3

Therefore, we have n = 7 and r = 3.

25.  We know that general term in the expansion of (1 + x)n is Tr+1 = nCrxr

Therefore Coefficient of xn in the expansion of (1 + x)2n is

From eq.(i) and eq.(ii) it is clear that coefficient of xn in the expansion of (1 + x)2n is twice the coefficient of xn in the expansion of (1 + x)2n-1.

26.  The general term in the binomial expansion of (x + a)is given by Tr+1 = nCrxn-rar

Assuming that x2 occur in the (r + 1)th term of expression (1 + x)m, we obtain

Tr+1 = mCr(1)m-r(x)mCr(x)r

Comparing the indices of x2 in term we get r = 2

so, coefficient of x2 is mC2. So

So, m = 4 or -3

Thus the positive value of m for which coefficient of x2 in expression (1 + x)m is 6 is 4.

27.   We have,

(a + b)n = nC0an + nC1an-1b + nC2an-2b2 + nC3an-3b3+…………+ nCnanbn

Squaring both sides of eq. (ii)

n2a2n-2b2 = (7290)2   …(v)

Dividing eq.(iv) by eq. (v),

∴ From eq. (i), a6 = 729

⇒ a6 = 36

⇒ a = 3

And from eq. (ii), 6 × 33 × b = 7290

⇒ b = 5

Therefore, n = 6, a = 3 and b = 5.

28.   We have,

(a + b)n = nC0an + nC1an-1b + nC2an-2b2 + nC3an-3b3+…………+ nCnanbn

∴ (3 + ax)9 = 9C039 + 9C138(ax) + 9C237(ax)2 + 9C336(ax)3+…………+ 9C9(ax)9

9C039 + 9C138ax + 9C237a2x2 + 9C336a3x3+…….+9C9a9x9

∴ Coefficient of x2 = 9C2(3)7 a2 and Coefficient of x3 = 9C3(3)6 a3

According to question,  9C2(3)7 a2 = 9C3(3)6 a3

29.   Using Binomial Theorem,

It is enough to consider only those terms in the product which involve x5 and they are

1(-21x5) + (12x) (35x4) + (60x2) (-35x3) + (160x3) (21x2) + (240x4) (-7x) + (192x5) (7x)

∴ Coefficient of x5 in the product

= {1 × (-21)} + {12 × 35} + {60 × (-35)} + {160 × 21} + {240 × (-7)} + {192 × 1}

= -21 + 420 – 2100 + 3360 – 1680 + 192 = 171.

30.  We have, an = [(a – b) + b]n

⇒ an – bn = k(a – b), where

k = [(a – b)n-1 + nC1(a – b)n-2 b + ………. + nCn-1bn-1] is an integer which shows that (a – b) is a factor of an – bn.

31.  First we will consider (a + b)6 – (a – b)6

Using Binomial Theorem, we have

32.   

33.  We have,

(1 + x)n = nC0 + nC1x + nC2x2 + nC3x3+…………+ nCnxn

Here (0.99)5 = (1 – 0.01)5 = [1 + (0.01)5] = 5C05C1 (0.01) – 5C2 (0.01)2   ….

= 1 – 5 × (0.01) – 10 × (0.01)2 ……  = 1 – 0.05 + 0.001 …….. = 0.001 – 0.05 = 0.951

34.  We have the general term of the binomial expansion

(a + b)n = nC0an + nC1an-1b + nC2an-2b2 + nC3an-3b3+…………+ nCnbn is Tr+1 = nCran-rbr

Also rth term from the end = (n – r + 2) term from the beginning

∴ 5th term from the beginning = T5 = T4+1 = nC4an-4band

5th term from the beginning = Tn-5+2 = Tn-3 = T(n-4)+1 = nCn-4a4bn-4

35.  We have, (1 + x)n = nC0 + nC1x + nC2x2 + nC3x3+…………+ nCnxn

36.  Here (3x2 – 2ax + 3a2)3 = [(3x – 2ax) + 3a2]3

= 3C (3x2 – 2ax)3 + 3C(3x2 – 2ax)2 (3a2) + 3C(3x2 – 2ax) (3a2)2 + 3C (3a2)3

= (3x2 – 2ax)3 + 3 × 3a2 (3x2 – 2ax)2 + 3 × 9a4 (3x2 – 2ax) + 27a6

= 27x6 – 8a3x3 – 54ax5 + 36a2x4 + 9a2 (9x4 + 4a2x2 – 12ax3) + 27a4 (3x2 – 2ax) + 27a6

= 27x6 – 8a3x3 – 54ax5 + 36a2x4 + 81a2x4 + 36a4x2 – 108a3x3 + 81a4x2 – 54a5x + 27a6

= 27x6 – 54ax5 + 117a2x4 – 116a3x3 + 117a4x2 – 54a5x + 27a3.

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