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NCERT Class XI Mathematics - Binomial Theorem - Solutions

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Question : 9 of 36
Marks: +1, -0
(99)5(99)^5
Solution:  
We have, 99 = 100 - 1
∴ (99)5(99)^5 = (100−1)5(100-1)^5 = [100+(−1)]5[100+(-1)]^5
= 5C0(100)5+5C1(100)4(−1){}^{5}C_{0}(100)^5 + {}^{5}C_{1}(100)^4(-1) + 5C2(100)3(−1)2+5C3(100)2(−1)3{}^{5}C_{2} (100)^3(-1)^2 + {}^{5}C_{3}(100)^2(-1)^3 + 5C4(100)(−1)4+5C5(−1)5{}^{5}C_{4}(100)(-1)^4 + {}^{5}C_{5}(-1)^5
= 10000000000 – 500000000 + 10000000 – 100000 + 500 – 1 = 9509900499.
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