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Question

1+22+322+423+........ to n terms.

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Solution

S=1+22+322+423+....n2n1S2=12+222+323+....n12n1+n2nSS2=1+12+122+123+....+12n1n2nS2=(112n)(112)n2nS2=2(2n1)2nn2nS=4(2n1)2n2n=2n+1n22n1

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