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.
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\\\\\n2=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]=\\\\\n(1-(-7)^{k+1})\/4 +2\\cdot (-7)^{k+1}=\\frac{1-(-7)^{k+1}+8\\cdot (-7)^{k+1}}{4}=\\\\\n=\\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.
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