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I am a first year Math student and I am looking at problem in my text book which does not have any answers and I have completely no idea how to do this paticular problem.

Show, using mathematical induction, that for all natural numbers n ≥ 3, $$ 4^2 + 4^3 + 4^4 + · · · + 4^n = \frac{4^2(4^{n-1} -1)}{3} $$

Base Case: n=3 $$4^2 + 4^3 = \frac{4^2(4^{n-1} -1)}{3}$$

I understand now

$$ 16 + 64 = 80 $$

Inductive step:

$$\frac{4^2(4^{n-1} -1)}{3}+4^{n+1}=\frac{4^2(4^n -1)}{3}\ .$$ what do I do after this?

Patrica Harris
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3 Answers3

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Hint. For the inductive step, you need to prove that $$\frac{4^2(4^{n-1} -1)}{3}+4^{n+1}=\frac{4^2(4^n -1)}{3}\ .$$

David
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But this is the well known formula for geometric sequence.

$a_1, a_1q, a_1q^2, \ldots, a_1q^{n-2}=a_1\frac{q^{n-1}-1}{q-1}$ with $a_1=4^2$ and $q=4$.

kmitov
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For n=3,
$4^2+4^3=\frac{4^2(4^2-1)}{3}$
Now for any n=k, assume that,
$4^2 + 4^3 + 4^4 + · · · + 4^k = \frac{4^2(4^{k-1} -1)}{3}$
Now, it has to be true for k+1,
$4^2+4^3+...+4^k+4^{k+1}=\frac{4^2(4^k-1)}{3}$
$\frac{4^2(4^{k-1} -1)}{3}+4^{k+1}=\frac{4^2(4^k-1)}{3}$
$3.4^{k+1}=4^2(4^k-1-4^{k-1}+1)$
Observe that both sides have been multiplied by $3$ and $4^2$ has been taken common.
$3.4^{k+1}=4^2(4^k-4^{k-1})$
$3.4^{k+1}=4^{k+2}-4^{k+1}$
$4.4^{k+1}=4^{k+2}$
And thus LHS=RHS.

Aditya Agarwal
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