The sum -k-1-6k-3k-2k3k-1-2k-1-

The correct option is D 2∑_k=1^∞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= ...

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Byju's Answer

Standard XII

Mathematics

Complex Numbers

The sum ∑k=1∞...

Question

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

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Solution

The correct option is B 2
$\sum_{k = 1}^{\infty} \frac{6^{k}}{(3^{k} - 2^{k}) (3^{k + 1} - 2^{k + 1})} = \frac{3^{k} - 2^{k}}{(3^{k} - 2^{k}) (3^{k + 1} - 2^{k + 1})} = \frac{3^{k}}{3^{k} - 2^{k}} - \frac{3^{k = 1}}{3^{k + 1} - 2^{k + 1}} \sum_{k = 1}^{\infty} \frac{3^{k}}{3^{k} - 2^{k}} - \frac{3^{k + 1}}{3^{k + 1} - 2^{k + 1}} = \frac{3}{3 - 2} - L t n \to \infty \frac{3^{2}}{3^{n} - 2^{n}} = 3 - 1 = 2$

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The sum ∑∞k=16k(3k−2k)(3k+1−2k+1)=

The correct option is B 2∑∞k=16k(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−Ltn→∞323n−2n=3−1=2

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