Is there a better way to find the sum of the below series than by traditional method (equating to constant say S and multiply with common ratio and subtract ) $S = 1-\frac{4}{3^1}+\frac{9}{3^2}-\frac{16}{3^3}+\frac{25}{3^4}-\frac{36}{3^5}+\cdots$
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See http://math.stackexchange.com/questions/593996/how-to-prove-sum-n-0-infty-fracn22n-6 Here $n$th term $$=3n^2\left(-\dfrac13\right)^n$$ – lab bhattacharjee Nov 30 '16 at 17:46
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Hint : General term of the series is given by $t_n=(-1)^{n+1}\cdot \frac {n^2}{3^{n-1}}$.
Now can you use the result of some common summations and evaluate $\sum_{n=1}^\infty t_n$?
SchrodingersCat
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In that case, the numerator will be given by $\frac {n (n+1)}{2}$. – SchrodingersCat Nov 30 '16 at 15:40
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You're welcome. In case you're unaware, if you like an answer, you can upvote it by clicking the up arrow on the left of the answer and accept it by clicking the 'tick' on the left of the answer. – SchrodingersCat Nov 30 '16 at 16:03