Sequence And Series Question 483
Question: Sum of n terms of series $ 12+16+24+40+….. $ will be
[UPSEAT 1999]
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:
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+… $ + upto n terms) - $ T_{n} $ or $ T_{n}=12+[4+8+16+…+\text{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}+….upto,n,\text{terms)} $ $ +(8+8+8+…… $ upto n terms) $ =\frac{2^{2}(2^{n}-1)}{2-1}+8n=4(2^{n}-1)+8n. $
BETA
