Question

Let C(n) be the constant term in the expansion of
(x + 4)n.
Prove by induction that
C(n) = 4n
for all n is in N.
(Induction on n.) The constant term of
(x + 4)1
is

4

Correct: Your answer is correct.
= 4

Incorrect: Your answer is incorrect.
.


Suppose as inductive hypothesis that the constant term of
(x + 4)k − 1
is

for some
k > 1.


Then
(x + 4)k = (x + 4)k − 1 ·




,
so its constant term is

· 4 =
,
as required.

Answer 1

Induction seems more work than necessary. The result follows from the binomial theorem,
\[\begin{align*}
(x+4)^n&=\sum_{k=0}^n\binom nkx^{n-k}4^k\\[1ex]
&=\underbrace{\binom nn4^n}_{C(n)}+\binom n{n-1}4^{n-1}x+\cdots+\binom n14x^{n-1}+\binom n0x^n
\end{align*}\]

Answer 2

If you're required to use induction, you have the base case when \(n=1\):
\[(x+4)^1=x+4\implies C(1)=4^1=4\]
Now assume \(C(k)=4^k\) is the constant term of \((x+4)^k\). Then
\[(x+4)^{k+1}=(x+4)(x+4)^k\]Distribute the terms to every term in the expansion of \((x+4)^k\), namely the constant term. Then
\[(x+4)C(k)=C(k)x+4C(k)\]with the constant term
\[4C(k)=4\times4^k=4^{k+1}\]as needed.