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Question

For the series S=1+1(1+3)(1+2)2+1(1+3+5)(1+2+3)2+1(1+3+5+7)(1+2+3+4)2+

A
9th term is 25
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B
7th term is 16
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C
Sum of first ten terms is 5054
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D
Sum of first ten terms is 505
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Solution

The correct option is C Sum of first ten terms is 5054
We have, S=1+1(1+3)(1+2)2+1(1+3+5)(1+2+3)2+1(1+3+5+7)(1+2+3+4)2+
So, the general term can be written as
Tr=1r2(1+2++r)2=1r2(r(r+1)2)2
=(r+1)24=r2+2r+14
T7=16
And, T9=25

S10=10r=1r2+210r=1r+10r=114
=14[10×11×216+2(10×112)+10]
=5054

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