The sum - k -1-6 k -3 k -2 k 3 k -1-2 k -1 is equalto

6^k/(3^k 2^k)(3^k+1 2^k+1)=3^k.2^k/(3^k 2^k)(3^k+1 2^k+1) =3^k/3^k 2^k 3^k+1/3^k+1 2^k+1 ∑_k=1^∞3^k/3^k 2^k 3^k+1/3^k+1 2^k+1=3/3 2 lim_n →∞3^n/3^n 2^n =3 1=2 ...

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Standard XII

Mathematics

Binomial Coefficients

The sum ∑ k =...

Question

The sum $\sum_{k = 1}^{\infty} \frac{6^{k}}{(3^{k} - 2^{k}) (3^{k + 1} - 2^{k + 1})}$ is equalto___

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Solution

$\frac{6^{k}}{(3^{k} - 2^{k}) (3^{k + 1} - 2^{k + 1})} = \frac{3^{k} {.2}^{k}}{(3^{k} - 2^{k}) (3^{k + 1} - 2^{k + 1})} = \frac{3^{k}}{3^{k} - 2^{k}} - \frac{3^{k + 1}}{3^{k + 1} - 2^{k + 1}} \sum_{k = 1}^{\infty} \frac{3^{k}}{3^{k} - 2^{k}} - \frac{3^{k + 1}}{3^{k + 1} - 2^{k + 1}} = \frac{3}{3 - 2} - {lim}_{n \to \infty} \frac{3^{n}}{3^{n} - 2^{n}} = 3 - 1 = 2$

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The sum ∑∞k=16k(3k−2k)(3k+1−2k+1) is equalto___

6k(3k−2k)(3k+1−2k+1)=3k.2k(3k−2k)(3k+1−2k+1)=3k3k−2k−3k+13k+1−2k+1∑∞k=13k3k−2k−3k+13k+1−2k+1=33−2−limn→∞3n3n−2n=3−1=2

The sum $\sum_{k = 1}^{\infty} \frac{6^{k}}{(3^{k} - 2^{k}) (3^{k + 1} - 2^{k + 1})}$ is equalto___