Question

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

(x + 9)n.
Prove by induction that
C(n) = 9n
for all n is in N.
(Induction on n.) The constant term of
(x + 9)1
is

= 9
.


Suppose as inductive hypothesis that the constant term of
(x + 9)k − 1
is

for some
k > 1.


Then
(x + 9)k = (x + 9)k − 1 ·




,
so its constant term is

· 9 =
,
as required.

Answer 1

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.