Sequence And Series Question 210
Question: $ 1+3+7+15+31+………. $ to $ n $ terms =
[IIT 1963]
Options:
A) $ {2^{n+1}}-n $
B) $ {2^{n+1}}-n-2 $
C) $ 2^{n}-n-2 $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Let $ T_{n} $ be the $ n^{th} $ term and $ S $ the sum upto $ n $ terms. $ S=1+3+7+15+31+……+T_{n} $ Again $ S=1+3+7+15+………..\text{ }+{T_{n-1}}+T_{n} $ Subtracting, we get $ 0=1+{ 2+4+8+…(T_{n}-{T_{n-1}}) }-T_{n} $
$ \therefore \ \ T_{n}=1+2+2^{2}+2^{3}+…..upto\ n\ terms $ $ =\frac{1(2^{n}-1)}{2-1}=2^{n}-1 $ Now $ S=\Sigma T_{n}=\Sigma 2^{n}-\Sigma 1 $ $ =(2+2^{2}+2^{3}+……+2^{n})-n $ $ =2( \frac{2^{n}-1}{2-1} )-n={2^{n+1}}-2-n $ . Aliter: $ 1+3+7+……+T_{n} $ $ =2-1+2^{2}-1+2^{3}-1+……….+2^{n}-1 $ $ =(2+2^{2}+……+2^{n})-n={2^{n+1}}-2-n $ . Trick: Check the options for $ n=1,\ 2 $ .
BETA
