Consider the sequence $A_n$$=\frac{81\pi }{2^{2n-2}}, \forall n\in \mathbb{N}$.
I'm supposed to prove by induction that the sum of the $n$ first terms of $A_n$ is given by $S_n$$=27\pi \left(4-2^{2-2n}\right)$. Here's what I've maneged to do:
Let $P(n)$: $81\pi +\frac{81\pi }{4}+\frac{81\pi }{16}+...+\frac{81\pi }{2^{2n-2}}=27\pi \left(4-2^{2-2n}\right)$
I. $n=1 \rightarrow 27\pi \left(4-2^{0}\right)=81\pi$ (True). Hence $P(1)$ is true.
II. By hypothesis we know that $81\pi +\frac{81\pi }{4}+\frac{81\pi }{16}+...+\frac{81\pi }{2^{2n-2}}=27\pi \left(4-2^{2-2n}\right)$. Then $P(n+1)=81\pi +\frac{81\pi }{4}+\frac{81\pi }{16}+...+\frac{81\pi }{2^{2n-2}} + \frac{81\pi }{2^n}=27\pi \left(4-2^{2-2n}\right)+\frac{81\pi }{2^n}$
I get stuck trying to put $27\pi \left(4-2^{2-2n}\right)+\frac{81\pi }{2^n}$ in the form $27\pi \left(4-2^{-2n}\right)$ (which is $P(n+1)$), required to complete the proof.
The last term in the sum for $P(n+1)$ is $\frac{81\pi}{2^{2(n+1)-2}}=\frac{81\pi}{2^{2n}}$, not $\frac{81\pi}{2^{n}}$. Then by induction hypothesis $$P(n+1)=P(n)+\frac{81\pi}{2^{2n}}=27\pi(4-2^{2-2n})+27\pi\frac{3}{2^{2n}}=27\pi\left(4-2^{2-2n}+3\cdot2^{-2n}\right),$$ from which point I am sure you can finish.