<s>[INST] Algebraic plateau in proof by induction 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.
 [/INST] 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.
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