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Pachacha [2.7K]
3 years ago
7

Let C(n) be the constant term in the expansion of

Mathematics
1 answer:
DochEvi [55]3 years ago
8 0
Presumably you meant to write

C(n)=9^n

For n=1, we have

(x+9)^1=x+9
C(1)=9^1=9

Suppose the claim holds for n=k, i.e. that

C(k)=9^k

Then for n=k+1, we have

(x+9)^{k+1}=(x+9)^k(x+9)=x(x+9)^k+9(x+9)^k

Every term in the expansion of the first term will have degree at least 1 (x^{k+1} at the most and 9^kx at the least), so we can safely ignore these terms.

This leaves us with

9(x+9)^k

We already know the constant term of the expansion here is C(k)=9^k. Multiplying by 9, we then are left with C(k+1)=9^{k+1}, proving the claim.
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