Binomial Theorem - Solutions

CBSE Class–11 Mathematics

NCERT Solutions
Chapter - 8 Binomial Theorem
Exercise 8.1

Expand each of the expression in Exercises 1 to 5.

Ans. Using Binomial Theorem,

$= 1 + 5 (- 2 x) + 10 (4 x^{2}) + 10 (- 8 x^{3}) + 5 (16 x^{4}) + (- 32 x^{5})$

= Ans.

Ans. Using Binomial Theorem,

${(\frac{2}{x} - \frac{x}{2})}^{5} =^{5} C_{0} {(\frac{2}{x})}^{5} +^{5} C_{1} {(\frac{2}{x})}^{4} {(\frac{- x}{2})}^{1} +^{5} C_{2} {(\frac{2}{x})}^{3} {(\frac{- x}{2})}^{2} +^{5} C_{3} {(\frac{2}{x})}^{2} {(\frac{- x}{2})}^{3} +^{5} C_{4} {(\frac{2}{x})}^{1} {(\frac{- x}{2})}^{4} +^{5} C_{5} {(\frac{- x}{2})}^{5}$

= $\frac{32}{x^{5}} + 5 (\frac{16}{x^{4}}) (\frac{- x}{2}) + 10 (\frac{8}{x^{3}}) ({\frac{x}{4}}^{2}) + 10 (\frac{4}{x^{2}}) (\frac{- x^{3}}{8}) + 5 (\frac{2}{x}) ({\frac{x}{16}}^{4}) + (\frac{- x^{5}}{32})$

Ans. Using Binomial Theorem,

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

Ans. Using Binomial Theorem,

= $x^{6} + 6 x^{5} \cdot \frac{1}{x} + 15 x^{4} \cdot \frac{1}{x^{2}} + 20 x^{3} \cdot \frac{1}{x^{3}} + 15 x^{2} \cdot \frac{1}{x^{4}} + 6 x \cdot \frac{1}{x^{5}} + \frac{1}{x^{6}}$

Using binomial theorem evaluate each of the following:

Ans. 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

Using Binomial Theorem,

= 1000000 – 120000 + 4800 – 64

= 1004800 – 120064 = 884736

Ans. 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

Using Binomial Theorem,

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

= 11040808032

Ans. 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

Using Binomial Theorem,

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

= 100000000 + 4000000 + 60000 + 400 + 1

= 104060401

Ans. 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

Using Binomial Theorem,

= $100^{5} + 5 (100)^{4} (- 1) + 10 (100)^{3} (1) + 10 (100)^{2} (- 1) + 5 (100) (1) + (- 1)$

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

= 9509900499

10. Using binomial theorem, indicate which number is larger or 1000.

Ans. We have $1.1 = 1 + 0.1$

Using Binomial Theorem,

= 1 + 10000 (0.1) + other positive numbers

= 1 + 1000 + other positive numbers

which is greater than 1000

11. Find Hence evaluate:

Ans. Given:

Using Binomial Theorem,

Putting and

= $8 \sqrt{6} [3 + 2] = 40 \sqrt{6}$ Ans.

12. Find Hence or otherwise evaluate

Ans. Given:

Using Binomial Theorem,

= =

Putting

= 2 [ 8 + 60 + 30 + 1] = = 198

13. Show that is divisible by 64 whenever is a positive integer.

Ans. We know that b is divisible by a( or a divides b) $\Rightarrow b = a k$ , k is an integer

Here we have to show that 64 divides

$\Rightarrow 9^{n + 1} - 8 n - 9 = 64 k$ ,k is an integer

We have

Using Binomial Theorem,we have

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

$9^{n + 1} =$

$9^{n + 1} - 8 n - 9 = 64 [^{n + 1} C_{2} +^{n + 1} C_{3} (8) + . . . . . . . . . . . . . . . . . . . . . . . . . +^{n + 1} C_{n + 1} {(8)}^{n - 1}]$

$\Rightarrow 9^{n + 1} - 8 n - 9 = 64 k$ ,where $k =^{n + 1} C_{2} +^{n + 1} C_{3} (8) + . . . . . . . . . . +^{n + 1} C_{n + 1} (8)^{n - 1}$ is an integer

which shows that is divisible by 64.

14. Prove that

Ans. L.H.S. =
But we have $(1 + x)^{n} =^{n} C_{0} +^{n} C_{1} x +^{n} C_{2} x^{2} +^{n} C_{3} x^{3} + . . . . . . . . . . . . . . . . . . . . . . . . . +^{n} C_{n} x^{n}$

$\sum_{r = 0}^{n} 3^{r}^{n} C_{r} =$
Hence proved