Sequence And Series Question 347
Question: Sum of n terms of series $ 12+16+24+40+… $ will be
Options:
A) $ 2(2^{n}-1)+8n $
B) $ 2(2^{n}-1)+6n $
C) $ 3(2^{n}-1)+8n $
D) $ 4(2^{n}-1)+8n $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Let nth term of series is $ T_{n} $ then $ S_{n}=12+16+24+40+….+T_{n} $ Again $ S_{n}=12+16+24+….+T_{n} $ On subtraction $ 0=(12+4+8+16+….+uptonterms),-T_{n} $ or $ T_{n}=12+[4+8+16+…+,upto(n-1)terms] $ $ =12+\frac{4({2^{n+1}}-1)}{2-1}={2^{n-1}}+8 $ On putting n = 1, 2, 3?? $ T_1=2^{2}+8,,T_2=2^{3}+8,,T_3=2^{4}+8…. $ etc. $ S_{n}=T_1+T_2+T_3+….+T_{n} $ $ =(2^{2}+2^{3}+2^{4}+…..uptonterms)+(8+8+8+…. $ $ uptonterms) $ $ =\frac{2^{2}(2^{n}-1)}{2-1}+8n=4(2^{n}-1)+8n $ .
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