# Class 11 NCERT Solutions- Chapter 8 Binomial Theorem – Exercise 8.1

Theorem 1:

(a+b)n nCk an-k bk

Here, the coefficients nCk are known as binomial coefficients.

Theorem 2:

(aâ€“b)n (-1)n nCk an-k bk

### Question 1. (1 â€“ 2x)5

Solution:

According to theorem 2, we have

a = 1

b = 2x

and, n = 5

So, (1 – 2x)5 = 5C0 (1)5 â€“ 5C1 (1)4 (2x)1 + 5C2 (1)3 (2x)2 â€“ 5C3 (1)2 (2x)3 + 5C4 (1)1 (2x)4 â€“ 5C5 (2x)5

= 1 â€“ 5 (2x) + 10 (4x)2 â€“ 10 (8x3) + 5 (16 x4) â€“ (32 x5)

= 1 â€“ 10x + 40x2 â€“ 80x3 + 80x4â€“ 32x5

### Question 2.

Solution:

According to theorem 2, we have

a =

b =

and, n = 5

So, = 5C0 ()5 â€“ 5C1 ()4 ()1 + 5C2 ()3 ()2 â€“ 5C3 ()2 ()3 + 5C4 ()1 ()4 â€“ 5C5 ()5

â€“ 5  + 10  â€“ 10  + 5  â€“

### Question 3. (2x â€“ 3)6

Solution:

According to theorem 2, we have

a = 2x

b = 3

and, n = 6

So, (2x â€“ 3)6 = 6C0 (2x)6 â€“ 6C1 (2x)5 (3)1 + 6C2 (2x)4 (3)2 â€“ 6C3 (2x)3 (3)3 + 6C4 (2x)2 (3)4 â€“ 6C5 (2x)1 (3)5 + 6C6 (3)6

= 64x6 â€“ 6(32x5)(3) + 15 (16x4) (9) â€“ 20 (8x3) (27) + 15 (4x2) (81) â€“ 6 (2x) (243) + 729

= 64x6 â€“ 576x5 + 2160x4 â€“ 4320x3 + 4860x2 â€“ 2916x + 729

### Question 4.

Solution:

According to theorem 1, we have

a =

b =

and, n = 5

So,  = 5C0 ()5 + 5C1 ()4 ()1 + 5C2 ()3 ()2 + 5C3 ()2 ()3 + 5C4 ()1 ()4 + 5C5 ()5

### Question 5.

Solution:

According to theorem 1, we have

a = x

b =

and, n = 6

So,  = 6C0 (x)6 + 6C1 (x)5 ()1 + 6C2 (x)4 ()2 + 6C3 (x)3 ()3 + 6C4 (x)2 ()4 + 6C5 (x)1 ()5 + 6C6 ()6

### Question 6. (96)3

Solution:

Given: (96)3

Here, 96 can be expressed as (100 – 4).

So, (96)3 = (100 â€“ 4)3

According to Theorem 2, we have

= 3C0 (100)3 â€“ 3C1 (100)2 (4) â€“ 3C2 (100) (4)2â€“ 3C3 (4)3

= (100)3 â€“ 3 (100)2 (4) + 3 (100) (4)2 â€“ (4)3

= 1000000 â€“ 120000 + 4800 â€“ 64

= 884736

### Question 7. (102)5

Solution:

Given: (102)5

Here, 102 can be expressed as (100 + 2).

So, here (102)5 = (100 + 2)5

According to Theorem 1, we have

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

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

= 10000000000 + 1000000000 + 40000000 + 80000 + 8000 + 32

= 11040808032

### Question 8. (101)4

Solution:

Given: (101)4

Here, 101 can be expressed as (100 + 1).

So, here (101)4 = (100 + 1)4

According to Theorem 1, we have

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

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

= 100000000 + 400000 + 60000 + 400 + 1

= 1040604001

### Question 9. (99)5

Solution:

Given: (99)5

Here, 99 can be expressed as (100 – 1).

So, here (99)5 = (100 â€“ 1)5

According to Theorem 2, we have

= 5C0 (100)5 â€“ 5C1 (100)4 (1) + 5C2 (100)3 (1)2 â€“ 5C3 (100)2 (1)3 + 5C4 (100) (1)4 â€“ 5C5 (1)5

= (100)5 â€“ 5 (100)4 + 10 (100)3 â€“ 10 (100)2 + 5 (100) â€“ 1

= 1000000000 â€“ 5000000000 + 10000000 â€“ 100000 + 500 â€“ 1

= 9509900499

### Question 10. Using Binomial Theorem, indicate which number is larger (1.1)10000 or 1000.

Solution:

Given: (1.1)10000

Here, 1.1 can be expressed as (1 + 0.1)

So, here (1.1)10000 = (1 + 0.1)10000

According to Theorem 1, we have

(1 + 0.1)10000 = 10000C0 (1)10000 + 10000C1 (1)9999 (0.1)1 + other positive terms

= 1 + 1000 + other positive terms

= 1100 + other positive terms

So, 1100 + other positive terms > 1000

Hence, proved (1.1)10000 > 1000

### Question 11. Find (a + b)4 â€“ (a â€“ b)4. Hence, evaluate (âˆš3 + âˆš2)4 â€“ (âˆš3 â€“ âˆš2)4.

Solution:

According to Theorem 1, we have

(a + b)4 = 4C0 a4 + 4C1 a3 b + 4C2 a2 b2 + 4C3 a b3 + 4C4 b4

According to Theorem 2, we have

(a â€“ b)4 = 4C0 a4 â€“ 4C1 a3 b + 4C2 a2 b2 â€“ 4C3 a b3 + 4C4 b4

Now, (a + b)4 â€“ (a â€“ b)4

= 4C0 a4 + 4C1 a3 b + 4C2 a2 b2 + 4C3 a b3 + 4C4 b4 â€“ [4C0 a4 â€“ 4C1 a3 b + 4C2 a2 b2 â€“ 4C3 a b3 + 4C4 b4]

= 2 (4C1 a3 b + 4C3 a b3)

= 2 (4a3 b + 4ab3)

= 8ab (a2 + b2)                    -(1)

Now, according to Equation(1), we get

a = âˆš3 and b = âˆš2

So, (âˆš3 + âˆš2)4 â€“ (âˆš3 â€“ âˆš2)4

= 8 Ã— âˆš3 Ã— âˆš2 ((âˆš3)2 + (âˆš2)2)

= 8 (âˆš6)(3 + 2)

= 40 âˆš6

### Question 12. Find (x + 1)6+ (x â€“ 1)6. Hence or otherwise evaluate (âˆš2 + 1)6 + (âˆš2 â€“ 1)6.

Solution:

According to Theorem 1, , we have

(x + 1)6 = 6C0 x6 + 6C1 x5 + 6C2 x4 + 6C3 x3 + 6C4 x2 + 6C5 x + 6C6

According to Theorem 2, , we have

(x â€“ 1)6 = 6C0 x6 â€“ 6C1 x5 + 6C2 x4 â€“ 6C3 x3 + 6C4 x2 â€“ 6C5 x + 6C6

Now, (x + 1)6 â€“ (x â€“ 1)6

= 6C0 x6 + 6C1 x5 + 6C2 x4 + 6C3 x3 + 6C4 x2 + 6C5 x + 6C6 â€“ [6C0 x6 â€“ 6C1 x5 + 6C2 x4 â€“ 6C3 x3 + 6C4 x2 â€“ 6C5 x + 6C6]

= 2 [6C0 x6 + 6C2 x4 + 6C4 x2 + 6C6]

= 2 [x6 + 15x4 + 15x2 + 1]                           -(1)

Now, According to Equation(1),

x = âˆš2

So, (âˆš2 + 1)6 â€“ (âˆš2 â€“ 1)6

= 2 [(âˆš2)6 + 15(âˆš2)4 + 15(âˆš2)2 + 1]

= 2 (8 + (15 Ã— 4) + (15 Ã— 2) + 1)

= 2 (8 + 60 + 30 + 1)

= 2 (99)

= 198

### Question 13. Show that 9n+1 â€“ 8n â€“ 9 is divisible by 64, whenever n is a positive integer.

Solution:

To Prove: 9n+1 â€“ 8n â€“ 9 = 64 k, where k is some natural number

According to Theorem 1, we have

For a = 1, b = 8 and m = n + 1 we get,

(1 + 8)n+1 = n + 1C0 + n + 1C1 (8) + n + 1C2 (8)2 + â€¦. +  n+1 C n+1 (8)n+1

9n+1 = 1 + (n + 1) 8 + 82 [n+1C2 + n+1C3 (8) + â€¦. + n+1 C n+1 (8)n+1]

9n+1 = 9 + 8n + 64 [n+1C2 + n+1C3 (8) + â€¦. + n+1 C n+1 (8)n+1]

9n+1 â€“ 8n â€“ 9 = 64 k

Where k, will be a natural number

Hence, proved 9n+1 â€“ 8n â€“ 9 is divisible by 64, whenever n is positive integer.

### Question 14. Prove that  3rnCr = 4n

Solution:

As, we know that According to Binomial Theorem,

nCk an-k bk = (a + b)n

By comparing Theorem 1 with question, we get

3r nCr = 4n

a + b = 4, k = r and b = 3

a = 1.

So,  nCr an-r br = (a+b)n

nCr 1n-r 3r = (1+3)n

nCr (1) 3r = 4n

nCr 3r = 4n

Hence, Proved

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