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(∑n=1∞5n16n)+(43+49+427+481+...)44(\sum\limits_{n=1}^\infty \frac{5 ^ n}{16 ^ n})+\frac{(\frac{4}{3}+\frac{4}{9}+\frac{4}{27}+\frac{4}{81}+...)}{44}(n=1∑∞16n5n)+44(34+94+274+814+...) can be written as ab\frac{a}{b}ba, where aaa and bbb are coprime positive integers. Find the value of ∑n=04an+bn\sum\limits_{n=0}^4 a^n+b^nn=0∑4an+bn.
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