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NCERT Class XI Mathematics - Binomial Theorem - Solutions

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Question : 30 of 36
Marks: +1, -0
If a and b are distinct integers, prove that a – b is a factor of an−bna^n - b^n, whenever n is a positive integer.
[Hint: write ana^n = (a−b+b)n(a - b + b)^n and expand]
Solution:  
We can write ana^n = (a−b+b)n(a - b + b)^n
Then ana^n = [(a−b)+b]n[(a-b)+b]^n
= nC0(a−b)n+nC1(a−b)n−1b{}^{n}C_{0}(a-b)^n + {}^{n}C_{1}(a-b)^{n-1}b + ... + nCn−1(a−b)bn−1{}^{n}C_{n-1}(a-b)b^{n-1} + nCnbn{}^{n}C_n b^n
⇒ an−bna^n - b^n = nC0(a−b)n+nC1(a−b)n−1b{}^{n}C_{0}(a-b)^n + {}^{n}C_{1}(a-b)^{n-1} b + ... + nCn−1(a−b)bn−1+bn−bn{}^{n}C_{n-1}(a-b)b^{n-1} + b^n - b^n
= (a - b) [nC0(a−b)n−1+nC1(a−b)n−2b{}^{n}C_{0}(a-b)^{n-1} + {}^{n}C_{1}(a-b)^{n-2}b + ... + nCn−1bn−1{}^{n}C_{n-1}b^{n-1}]
= (a – b) (some integer)
[nC0,nC1,nC2{}^{n}C_{0}, {}^{n}C_{1}, {}^{n}C_{2} , ... , nCn−1{}^{n}C_{n-1} , are integers & also all non negative powers of a – b and b are integers]
Hence, a – b is a factor of an−bna^n - b^n.
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