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Question

The sum of the following series 1+6+9(12+22+32)7+12(12+22+32+42)9+15(12+22+....+52)11+..... up to 15 terms is:

A
7820
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B
7830
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C
7520
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D
7510
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Solution

The correct option is A 7820
Tn=(3+(n1)×3)(12+22+...+n2)(2n+1)

Tn=3n.n(n+1)(2n+1)62n+1=n2(n+1)2

S15=1215n=1(n3+n2)=12(15(15+1)2)2+15×16×316

=7820

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