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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
7510
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B
7820
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C
7830
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D
7520
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Solution

The correct option is B 7820
S=1+6+9(12+22+32)7+12(12+22+32+42)9 +15(12+22+...+52)11+... up to 15 terms

S=33(12)+65(12+22)+129(12+22+32+42)+...+ up to 15 terms

Tn=3n(12+22+...+n2)(2n+1)

=3nn(n+1)(2n+1)6(2n+1)

Tn=12(n2+n3)

Sn=12[n2+n3]

Sn=12[n(n+1)(2n+1)6+n2(n+1)24]

Sn=12n(n+1)[2n+16+n(n+1)4]

S15=12×15(15+1)[316+15×164]

=12×15×16[62+72012]

S15=7820

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