Question #142108
Find the coefficient of x^7 y^9 in the expansion of (4x+ 5y)^16.
1
Expert's answer
2020-11-17T17:01:42-0500

Expansion of (a+b)n(a+b)^n gives us (n+1)(n+1) terms which are given by binomial expansion (nr)a(nr)br\dbinom{n}{r}a^{(n-r)}b^r , where rr ranges from nn to 0.

Note that powers of aa and bb add up to nn and in the given problem this n=7+9=16n=7+9=16.

In (4x+5y)16(4x+5y)^{16} , we need coefficient of x7y9x^7y^9 , we have 7th7^{th} power of xx and as such r=167=9r=16-7=9

and as such the desired coefficient of x7y9x^7y^9 is given by

(169)(4x)(169)(5y)9=16!9!(169)!(4x)7(5y)9=1144016384x71953125y9=3.6608E14x7y9\dbinom{16}{9}(4x)^{(16-9)}(5y)^9=\dfrac{16!}{9!(16-9)!}(4x)^7(5y)^9=11440*16384x^7*1953125y^9= 3.6608E14x^7y^9

So6 the coefficient is 3.6608E14

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