# Prove for all integers n such that n ≥ 3, $4^3 + 4^4 + 4^5 … 4^n = \frac{4(4^n - 16)}{3}$

I am trying to prove this using mathematical induction, but I'm lost once I get to comparing the two sides of the equation.

Proposition:

For all integers n such that n ≥ 3,

$4^3 + 4^4 + 4^5 … 4^n = \frac{4(4^n - 16)}{3}$

Proof:

Let the property P(n) be the equation

$P(n) = 4^3 + 4^4 + 4^5 … 4^n = \frac{4(4^n - 16)}{3}$

Show that P(3) is true:

$4^3 = \frac{4(4^3 - 16)}{3}$

64 = 64, thus P(3) is true

Show that for all integers where n ≥ 3, if P(k) is true, then P(k + 1) is also true:

Suppose that P(k) is true for some particular but arbitrary integer where k ≥ 3. Suppose that k is any integer where k ≥ 3 such that:

$4^3 + 4^4 + 4^5 … 4^k = \frac{4(4^k - 16)}{3}$

We must show that P(k + 1) is true. That is, we must show that:

$4^3 + 4^4 + 4^5 … 4^{k + 1} = \frac{4(4^{k + 1} - 16)}{3}$

The left hand side is:

$4^3 + 4^4 + 4^5 … 4^{k + 1}$

$4^3 + 4^4 + 4^5 … 4k + 4^{k + 1}$

$4^3 + 4^4 + 4^5 … \frac{4(4^k - 16)}{3} + 4^{k + 1}$

• You've made a mistake on the left hand side; you've substituted the P(k) hypothesis for $4^k$, instead of substituting it for the entire sum $1+...+4^k$. Feb 4, 2016 at 22:35
• The left hand side for $P(k+1)$ will end up being $P(k) +4^{k+1}$. You could potentially substitute in your closed form for $P(k)$ then, which may help.
– Mark
Feb 4, 2016 at 22:36

Using the induction hypothesis, the last line you wrote should be $\frac{4(4^k - 16)}{3} + 4^{k + 1}$. Then: \begin{align*} \frac{4(4^k - 16)}{3} + 4^{k + 1} &= \frac{(4^{k+1} - 4\cdot16)}{3} + \frac {3\cdot4^{k + 1}}3 \\ &= \frac{4^{k+1} - 4\cdot16+3\cdot4^{k + 1}}3 \\ &= \frac{4\cdot 4^{k+1} - 4\cdot16}3 \\ &= \frac{4\cdot (4^{k+1} - 16)}3 \end{align*}