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Algebra Level 3

∑ n = 0 ∞ 9 − n + 2 4 n + 1 = A B \displaystyle \sum^{\infty}_{n=0}9^{-n+2}4^{n+1}=\frac{A}{B}

Find A + B A+B where A A and B B are positive co-prime integers.


The answer is 2921.

Akshat Sharda by Akshat Sharda , 21, India

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1 solution

Chew-Seong Cheong
Nov 27, 2015

∑ n = 0 ∞ 9 − n + 2 4 n + 1 = ∑ n = 0 ∞ ( 9 2 ) ( 4 1 ) 9 − n 4 n = 324 ∑ n = 0 ∞ ( 4 9 ) n = 324 1 − 4 9 = 2916 5 \begin{aligned} \sum^\infty_{n=0} 9^{-n+2}4^{n+1} & = \sum^\infty_{n=0} (9^2)(4^1)9^{-n}4^{n} \\ & = 324 \sum^\infty_{n=0} \left(\frac{4}{9}\right)^n \\ & = \frac{324}{1-\frac{4}{9}} \\ & = \frac{2916}{5} \end{aligned}

⇒ A + B = 2916 + 5 = 2921 \Rightarrow A+B = 2916+5 = \boxed{2921}

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