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Question

Find the sum of the n terms of the series 1+3+7+15+31+n terms.


A

2(n+1)3n

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B

2(n+1)2+n

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C

2(n+1)+3+n

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D

2(n+1)2n

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Solution

The correct option is D

2(n+1)2n


We see that the differences between successive a term is 2,4,8,16,

This is a GP.
Sn=1+3+7+15+31+Tn1+Tn
(1)
Sn=1+3+7+15+Tn1+Tn
(2)

[Now, we write the Sn in such a way that 1st term of equation 2 comes under 2nd term of equation 1]

Then, Subtracting both, we get,
0=1+2+4+8+16++(Tn+Tn1)Tn
Tn=1+2+4+8+16+n terms
we get, Tn = 1.(2n1)21
Hence sum of the n terms
Sn=tn=ni=1(2i1)
Sn=2n1=2(n+1)2n

Aliter:
§n=1+3+7+15++nthterm=(211+221+231++2n1=21+22+23++2nn=2×12n12n=2+2n+1n


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