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Using binomial theorem, evaluate the following: (99)5

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Question

Using binomial theorem, evaluate the following:

(99)5

Sum
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Solution

99 can be written as the sum or difference of two numbers whose powers are easier to calculate and then, Binomial Theorem can be applied.

It can be written that, 99 = 100 – 1

∴ `(99)^5 = (100 - 1)^5`

= `""^5C_0  (100)^5  +  ^5C_1  xx  (100)^4  xx  (- 1)  +  ^5C_2  xx  (100)^3  xx  (- 1)^2  +  ^5C_3  xx  (100)^2  xx  (-  1)^3  +  ^5C_4  xx  (100)  xx  (-4)^4  +  (-1)^5`

= 10000000000 – 5 x 100000000 + 10 x 1000000 – 10 x 10000 + 5 x 100 – 1

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

= 9509900499

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Chapter 7: Binomial Theorem - EXERCISE 7.1 [Page 133]

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NCERT Mathematics [English] Class 11
Chapter 7 Binomial Theorem
EXERCISE 7.1 | Q 9. | Page 133

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