tonqa13or
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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.

Respuesta :

LammettHash
Presumably you meant to write

[tex]C(n)=9^n[/tex]

For [tex]n=1[/tex], we have

[tex](x+9)^1=x+9[/tex]
[tex]C(1)=9^1=9[/tex]

Suppose the claim holds for [tex]n=k[/tex], i.e. that

[tex]C(k)=9^k[/tex]

Then for [tex]n=k+1[/tex], we have

[tex](x+9)^{k+1}=(x+9)^k(x+9)=x(x+9)^k+9(x+9)^k[/tex]

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

This leaves us with

[tex]9(x+9)^k[/tex]

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

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