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

Theorem 1:(a+b)

^{n}=^{n}C_{k}a^{n-k}b^{k}Here, the coefficients

^{n}C_{k }are known as binomial coefficients.

Theorem 2:(aâ€“b)

^{n}= (-1)^{n}^{n}C_{k}a^{n-k}b^{k}

### Expand each of the expressions in Exercises 1 to 5.

### Question 1. (1 â€“ 2x)^{5}

**Solution:**

According to theorem 2, we have

a = 1

b = 2x

and, n = 5

So, (1 – 2x)

^{5}=^{5}C_{0}(1)^{5}â€“^{5}C_{1}(1)^{4}(2x)^{1}+^{5}C_{2}(1)^{3}(2x)^{2}â€“^{5}C_{3}(1)^{2}(2x)^{3}+^{5}C_{4}(1)^{1}(2x)^{4}â€“^{5}C_{5}(2x)^{5}= 1 â€“ 5 (2x) + 10 (4x)

^{2}â€“ 10 (8x^{3}) + 5 (16 x^{4}) â€“ (32 x^{5})= 1 â€“ 10x + 40x

^{2}â€“ 80x^{3}+ 80x^{4}â€“ 32x^{5}

### Question 2.

**Solution:**

According to theorem 2, we have

a =

b =

and, n = 5

So, =

^{5}C_{0}()^{5}â€“^{5}C_{1}()^{4}()^{1}+^{5}C_{2}()^{3}()^{2}â€“^{5}C_{3}()^{2}()^{3}+^{5}C_{4}()^{1}()^{4}â€“^{5}C_{5}()^{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}=^{6}C_{0}(2x)^{6}â€“^{6}C_{1}(2x)^{5}(3)^{1}+^{6}C_{2}(2x)^{4}(3)^{2}â€“^{6}C_{3}(2x)^{3}(3)^{3}+^{6}C_{4}(2x)^{2}(3)^{4}â€“^{6}C_{5}(2x)^{1}(3)^{5}+^{6}C_{6}(3)^{6}= 64x

^{6}â€“ 6(32x^{5})(3) + 15 (16x^{4}) (9) â€“ 20 (8x^{3}) (27) + 15 (4x^{2}) (81) â€“ 6 (2x) (243) + 729= 64x

^{6}â€“ 576x^{5}+ 2160x^{4}â€“ 4320x^{3}+ 4860x^{2}â€“ 2916x + 729

### Question 4.

**Solution:**

According to theorem 1, we have

a =

b =

and, n = 5

So, =

^{5}C_{0}()^{5}+^{5}C_{1}()^{4}()^{1}+^{5}C_{2}()^{3}()^{2}+^{5}C_{3}()^{2}()^{3}+^{5}C_{4}()^{1}()^{4}+^{5}C_{5}()^{5}=

=

### Question 5.

**Solution:**

According to theorem 1, we have

a = x

b =

and, n = 6

So, =

^{6}C_{0}(x)^{6}+^{6}C_{1}(x)^{5}()^{1}+^{6}C_{2}(x)^{4}()^{2}+^{6}C_{3}(x)^{3}()^{3}+^{6}C_{4}(x)^{2}()^{4}+^{6}C_{5}(x)^{1}()^{5}+^{6}C_{6}()^{6}=

=

### Using the binomial theorem, evaluate each of the following:

### 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

=

^{3}C_{0}(100)^{3}â€“^{3}C_{1}(100)^{2}(4) â€“^{3}C_{2}(100) (4)^{2}â€“^{3}C_{3}(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

=

^{5}C_{0}(100)^{5}+^{5}C_{1}(100)^{4}(2) +^{5}C_{2}(100)^{3}(2)2 +^{5}C_{3}(100)^{2}(2)3 +^{5}C_{4}(100) (2)^{4}+^{5}C_{5}(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

=

^{4}C_{0}(100)^{4}+^{4}C_{1}(100)^{3}(1) +^{4}C_{2}(100)^{2}(1)^{2}+^{4}C_{3}(100) (1)^{2 }+^{4}C_{4}(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

=

^{5}C_{0}(100)^{5}â€“^{5}C_{1}(100)^{4}(1) +^{5}C_{2}(100)^{3}(1)^{2}â€“^{5}C_{3}(100)^{2}(1)^{3}+^{5}C_{4}(100) (1)^{4}â€“^{5}C_{5}(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}=^{10000}C_{0}(1)^{10000}+^{10000}C_{1}(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}=^{4}C_{0}a^{4}+^{4}C_{1}a^{3}b +^{4}C_{2}a^{2}b^{2}+^{4}C_{3}a b^{3}+^{4}C_{4}b^{4}According to Theorem 2, we have

(a â€“ b)

^{4}=^{4}C_{0}a^{4}â€“^{4}C_{1}a^{3}b +^{4}C_{2}a^{2}b^{2}â€“^{4}C_{3}a b^{3}+^{4}C_{4}b^{4}Now, (a + b)

^{4}â€“ (a â€“ b)^{4}=

^{4}C_{0}a^{4}+^{4}C_{1}a^{3}b +^{4}C_{2}a^{2}b^{2}+^{4}C_{3}a b^{3}+^{4}C_{4}b^{4}â€“ [^{4}C_{0}a^{4}â€“^{4}C_{1}a^{3}b +^{4}C_{2}a^{2}b^{2}â€“^{4}C_{3}a b^{3}+^{4}C_{4}b^{4}]= 2 (

^{4}C_{1}a^{3}b +^{4}C_{3}a b^{3})= 2 (4a

^{3}b + 4ab^{3})= 8ab (a

^{2}+ b^{2})-(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}=^{6}C_{0}x^{6}+^{6}C_{1}x^{5}+^{6}C_{2}x^{4}+^{6}C_{3}x^{3}+^{6}C_{4}x^{2}+^{6}C_{5}x +^{6}C_{6}According to Theorem 2, , we have

(x â€“ 1)

^{6}=^{6}C_{0}x^{6}â€“^{6}C_{1}x^{5}+^{6}C_{2}x^{4}â€“^{6}C_{3}x^{3}+^{ 6}C_{4}x^{2}â€“^{6}C_{5}x +^{6}C_{6}Now, (x + 1)

^{6}â€“ (x â€“ 1)^{6}=

^{6}C_{0}x^{6}+^{6}C_{1}x^{5}+^{6}C_{2}x^{4}+^{6}C_{3}x^{3}+^{6}C_{4}x^{2}+^{6}C_{5}x +^{6}C_{6}â€“ [^{6}C_{0}x^{6}â€“^{6}C_{1}x^{5}+^{6}C_{2}x^{4}â€“^{6}C_{3}x^{3}+^{6}C_{4}x_{2}â€“^{6}C_{5}x +^{6}C_{6}]= 2 [

^{6}C_{0}x^{6}+^{6}C_{2}x^{4}+^{6}C_{4}x^{2}+^{6}C_{6}]= 2 [x

^{6}+ 15x^{4}+ 15x^{2}+ 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 9^{n+1} â€“ 8n â€“ 9 is divisible by 64, whenever n is a positive integer.

**Solution:**

To Prove: 9

^{n+1}â€“ 8n â€“ 9 = 64 k, where k is some natural numberAccording to Theorem 1, we have

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

(1 + 8)

^{n+1}= n +^{1}C_{0}+ n +^{1}C_{1}(8) + n +^{1}C_{2}(8)^{2}+ â€¦. +^{n+1}C_{n+1}(8)^{n+1}9

^{n+1}= 1 + (n + 1) 8 + 8^{2}[^{n+1}C_{2}+^{n+1}C_{3}(8) + â€¦. +^{n+1}C_{n+1}(8)^{n+1}]9

^{n+1}= 9 + 8n + 64 [^{n+1}C_{2}+^{n+1}C_{3}(8) + â€¦. +^{n+1}C_{n+1}(8)^{n+1}]9

^{n+1}â€“ 8n â€“ 9 = 64 kWhere k, will be a natural number

Hence, proved 9

^{n+1}â€“ 8n â€“ 9 is divisible by 64, whenever n is positive integer.

### Question 14. Prove that 3^{r} ^{n}C_{r} = 4^{n}

**Solution:**

As, we know that According to Binomial Theorem,

^{n}C_{k}a^{n-k}b^{k}= (a + b)^{n}By comparing Theorem 1 with question, we get

3

^{r}^{n}C_{r}= 4^{n}a + b = 4, k = r and b = 3

a = 1.

So,

^{n}C_{r}a^{n-r}b^{r}= (a+b)^{n}

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

^{n}C_{r}(1) 3^{r}= 4^{n}

^{n}C_{r}3_{r}= 4^{n}Hence, Proved

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