NIMCET 2011 Question Paper with solutions

Question : 7 of 120

Marks: +1, -0

The sum of

11^{2}+12^{2}+. . .+30^{2}

Solution:

∴ 11^{2}+12^{2}+. . .+30^{2}

=(1^{2}+z^{2}+. . .+30^{2})-(1^{2}+z^{2}+. . .+10^{2})

[ ∵ Σ n^{2}= {n(n+1)(2 n+1)}/{6}]

= {30(30+1)(60+1)}/{6}-[\;{10 ×(10+1)(20+1)}/{6}]

= {56730}/{6}- {2310}/{6}= {54420}/{6}=9070

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