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= ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Complex Numbers

The sum ∑k=1∞...

Question

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

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

Suggest Corrections

Similar questions

Q. The sum

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

is equalto___

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

is equal to

lim n \to \infty \frac{2^{k} + 4^{k} + 6^{k} + . . . + {(2 n)}^{k}}{n^{k + 1}}, k \neq 1,

is equal to

Q. Solve:

l i m n \to θ (\frac{1^{k} + 2^{k} + 3^{k} + . . . . . k}{n^{k + 1}}) = (k > - 1)

Q. Find $\sum_{k=0}^{\infty}12(\dfrac{2}{3})^k - \sum_{\infty}^{k=0}18(\dfrac{-1}{2})^k =$

Join BYJU'S Learning Program

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