In the expansion of (1+a)m+n prove that coefficients of am and an are equal.

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Filo · Accepted Answer

It is known that (r+1)th term (Tr+1​) in the binomial expansion of (a+b)n is given by Tr+1​=nCr​an−1br Assuming that am occurs in the (r+1)th term of the expansion (1+a)m+n we obtain Tr+1​=m+nCr​(1)m+n−r(a)r=m+nCr​ar Comparing the indices of a in am and in Tr+1​ we obtain r=mTherefore the coefficient of am ism+nCm​= m!(m+n−m)!(m+n)!​=m!n!(m+n)!​....(1)Assuming that an occurs in the (k+1)th term of expansion (1+a)m+n, we obtain Tk+1​=m+nCk​(1)m+n−k(a) k=m+nCk​akComparing the indices of a in an and in Tk+1​ we obtain k=nTherefore the coefficient of an ism+nCn​= n!(m+n−n)!(m+n)!​=n!m!(m+n)!​....(2) Thus from 1 and 2 it can be observed that the coefficients of am and an in the expansion of (1+a)m+n are equal.

Question Text	In the expansion of $(1 + a)^{m + n}$ prove that coefficients of $a^{m}$ and $a^{n}$ are equal.
Answer Type	Text solution:1
Upvotes	150

In the expansion of $(1 + a)^{m + n}$ prove that coefficients of $a^{m}$ and $a^{n}$ are equal.

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