Sequence And Series Question 388
Question: The sum of the series $ \frac{2}{3}+\frac{8}{9}+\frac{26}{27}+\frac{80}{81}+….n $ terms is:
Options:
A) $ n-\frac{1}{2}(3^{n}-1) $
B) $ n+\frac{1}{2}(3^{n}-1) $
C) $ n-\frac{1}{2}(1-{3^{-n}}) $
D) $ n+\frac{1}{2}({3^{-n}}-1) $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Taking the sequence $ 3,9,27,81,….. $ Its nth term $ =3,{{(3)}^{n-1}}=3^{n} $ Also take the sequence $ 2,8,26,80….. $ or $ (3-1), $ $ (9-1),,(27-1),(81-1),…… $ Its nth term $ =3^{n}-1 $ Hence, nth term of the sequence $ \frac{2}{3}+\frac{8}{9}+\frac{26}{27}+\frac{80}{81}+……. $ is $ \frac{3^{n}-1}{3^{n}} $ or $ 1-{3^{-n}} $ Now the sum $ (S_{n})=\Sigma (1-{3^{-n}}) $ $ =n-({3^{-1}}+{3^{-2}}+….+{3^{-n}}) $ $ =n-\frac{{3^{-1}}{1-{{({3^{-1}})}^{n}}}}{1-{3^{-1}}}=n-\frac{1}{2}(1-{3^{-n}}) $ $ =n+\frac{1}{2}({3^{-n}}-1). $
BETA
