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If `sum_(k=1)^(n)k=210,` find the value of `sum_(k=1)^(n)k^(2).` |
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Answer» Correct Answer - 2870 `(1+2+3+...+n)=210rArr (1)/(2)n(n+1)=210rArr n^(2)+n-420=0` `rArr (n+21)(n-20)=0 rArr n=20.` `therefore (1^(2)+2^(2)+3^(2)+...+n^(2))=(1)/(6)n(n+1)(2n+1), " where " n =20` `=((1)/(6)xx20xx21xx41)=2870.` |
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