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Permutations And Combinations

Question
CBSEENMA11013179

Find the sum to n terms of the following series:

52 + 62 + 72 + ....+ 202

Solution
space space space space space space space space space 5 squared plus 6 squared plus.... plus 20 squared space equals space left parenthesis 1 squared plus 2 squared plus 3 squared plus.... plus 20 squared right parenthesis space minus space left parenthesis 1 squared plus 2 squared plus 3 squared plus 4 squared right parenthesis
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                                    = space space space fraction numerator left parenthesis 20 right parenthesis space left parenthesis 21 right parenthesis space left parenthesis 41 right parenthesis over denominator 6 end fraction minus fraction numerator left parenthesis 4 right parenthesis space left parenthesis 5 right parenthesis space left parenthesis 9 right parenthesis over denominator 6 end fraction           open square brackets use space sum straight n squared space equals space fraction numerator straight n left parenthesis straight n plus 1 right parenthesis left parenthesis 2 straight n plus 1 right parenthesis over denominator 6 end fraction close square brackets

                                    = 2870 - 30 = 2840

Some More Questions From Permutations and Combinations Chapter

Determine K, so that K + 2, 4K – 6 and 3K – 2 are three consecutive terms of an A.P.