If A = sum _n = 1^infinity (1 / (3 + (-1)^n)^n) and B = sum _n = 1^infinity ((-1)^n / (3 + (-1)^n)^n) then (A / B) is equal to:

We have A = (1 / 2) + frac14^2 + frac12^3 + frac14^4 + ... ... and B = (- 1 / 2) + frac14^2 + frac- 12^3 + frac14^4 + ... ... \ Therefore A = ( (1 / 2) + frac12^3 + frac12^5 + ... ..infinity ) + ( frac14^2 + frac14^4 + ... .infinity )\ = ( frac(1 / 2)1 - (1 / 4) ) + ( fracfrac14^21 - frac14^2 ) = (11 / 15) Similarl B = frac(- 1 / 2)1 - (1 / 4) + fracfrac14^21 - frac14^2 = - (9 / 15) \ Therefore (A / B) = (11 / 15) x ( (- 15 / 9) ) = (- 11 / 9)

If A = \sum_{n = 1}^{\infty}\mspace{2mu}\frac{1}{( ... | Sequence & Series (Mathematics)

We have $A = \frac{1}{2} + \frac{1}{4^{2}} + \frac{1}{2^{3}} + \frac{1}{4^{4}} + \ldots\ldots\\$ and $B = \frac{- 1}{2} + \frac{1}{4^{2}} + \frac{- 1}{2^{3}} + \frac{1}{4^{4}} + \ldots\ldots$ $\\ \therefore A = \left( \frac{1}{2} + \frac{1}{2^{3}} + \frac{1}{2^{5}} + \ldots..\infty \right) + \left( \frac{1}{4^{2}} + \frac{1}{4^{4}} + \ldots.\infty \right)\\$ $= \left( \frac{\frac{1}{2}}{1 - \frac{1}{4}} \right) + \left( \frac{\frac{1}{4^{2}}}{1 - \frac{1}{4^{2}}} \right) \\= \frac{11}{15} \\$ Similarl $B = \frac{\frac{- 1}{2}}{1 - \frac{1}{4}} + \frac{\frac{1}{4^{2}}}{1 - \frac{1}{4^{2}}} \\= - \frac{9}{15}$ $\\ \therefore\frac{A}{B} = \frac{11}{15} \times \left( \frac{- 15}{9} \right) \\= \frac{- 11}{9}$

Explanation

Let a_{1},a_{2},,a_{n} be a given A.P. whose common difference is an integer and S_{n} = a_{1} + a_{2} + + a_{n}. If a_{1} = 1,a_{n} = 300 and 15 \leq n \leq 50, then the ordered pair ( S_{n - 4},a_{n - 4} ) is equal to

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