Prove that 2 − 2 · 7 + 2 · 72 −· · ·+2(−7)n = (1 – ((−7)n+1)/4 whenever n is a nonnegative integer

We use math induction method to prove.

1. Base of induction.

n=0.

Left part $2\cdot(-7)^0=2\cdot 1=2;$

Right part $(1-(-7)^{n+1})/4=(1-(-7)^1)/4=(1-(-7))/4=\frac{8}{4}=2\\ 2=2$

Base of induction is verified.

2 Inductive step.

Let we make a guess that the statement is true for n=k, or

$S(k)=2-2\cdot 7+...+2\cdot(-7)^k=(1-(-7)^{k+1})/4$

Relying on this guess we must prove the identity for n=k+1.

We have:

S(k+1)=S(k)+2\cdot (-7)^{k+1}=\\ [S(k)=(1-(-7)^{k+1})/4 \space by \space assumption]=\\ (1-(-7)^{k+1})/4 +2\cdot (-7)^{k+1}=\frac{1-(-7)^{k+1}+8\cdot (-7)^{k+1}}{4}=\\ =\frac{1-(-7)^{k+1}+(-7)^{k+1}-(-7)\cdot (-7)^{k+1}}{4}=\frac{1-(-7)^{k+2}}{4}

So statement is true for n=k+1 and Inductive step is verified.

Thus given identity is proved by induction.