$ \sum_{k=1}^n {\delta}_k= 3^n-1$
then $\sum_{k=1}^\infty \frac1{\delta_k}= $
I tried writing the reciprocal terms and taking lcm but it isnt feasible since there are infinite terms
$ \sum_{k=1}^n {\delta}_k= 3^n-1$
then $\sum_{k=1}^\infty \frac1{\delta_k}= $
I tried writing the reciprocal terms and taking lcm but it isnt feasible since there are infinite terms
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