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Question

The sum of first 9 terms of the series 131+13+231+3+13+23+331+3+5+.... is

A
71
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B
96
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C
142
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D
192
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Solution

The correct option is B 96
rth term of the given series is, Tr=13+23+33+.......+r31+3+5+.....+2r1

[By using ni=1k3=13+23+33++n3=n2(n+1)24 and Sn=n2[2a+(n1)d] we have

Tr=(r(r+1)2)2r2(2+(r1)2)

=14(r+1)2

=14(r2+2r+1)

Thus sum of n term is,

Tr=14[r2+2r+1]

=14(n(n+1)(2n+1)6+n(n+1)+n)

=14(n(n+1)(2n+1)6+212n(n+1)+n)

Putting n=9 we get required sum

=14(9(10)(19)6+9(10)+9)

=96

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