# how can we express 985 & 1025

We have,

985 and 1025

985 = (100 -2)5 and 1025 = (100 + 2)5

i)

985 = (100 - 2)5 can be expand by using binomial theorem as:

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

[Using: (x  y)n = nC0 xn  nC1 xn−1y + nC2 xn−2y2 − …… + (−1)n nCnyn]

(100 - 2)5 = 1.(100)5  5. (100)4(2) + 10(100)3(4) − 10(100)2(8) + 5(100)1(16) - 1(100)5−5(32)

⇒ (100 - 2)5 = 1010  10. (10)8 + 40(10)6 − 80(10)4 + 80(10)2 - (32)

⇒ (100 - 2)5 = 1010  109 + 40(10)6 − 80(10)4 + 80(10)2 - (32)

⇒ (100 - 2)5 = 1010  109 + 40(10)6 − 80(10)4 + 80(10)2 - (32)

⇒ (100 - 2)5 =  9039207968

ii)

1025 = (100 + 2)5 can be expand by using binomial theorem as:

(100 + 2)5 = 5C01005 + 5C11005−1(2) + 5C2(100)5−2(2)2 + 5C3(100)5−3(2)3 + 5C4(100)5−4(2)4 + 5C5(100)5−5(2)5

[Using: (a + b)n = nC0an + nC1an−1b +nC2an−2b2 + …… + nCn−1abn−1 + nCnbn]

⇒ (100 + 2)5 = 1. 1005 + 5. 1004 (2) + 10(100)3(4) + 10 (100)2(8) + 5(100)(16) + 1(100)0(32)

⇒ (100 + 2)5 = 1005 + 10. 1004  + 40(100)3 + 80(100)2 + 80(100) + 32

⇒ (100 + 2)5 = 1010 + 109  + 40.106 + 80.104 + 8000 + 32

⇒ (100 + 2)5 = 1010 + 109  + 40.106 + 80.104 + 8000 + 32

⇒ (100 + 2)5 = 11040808032

Hope you get it!!

• 0

(100-2)   and (100+2)5

• 2

but to expand it and get answer

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how to expand it

• 0

use binomial theorem

• 1

use binomial theoram as (100-2)5 & (100+2)5

• 0
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