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Permutations And Combinations

Question
CBSEENMA11013175

Sum of the series n . 1 + (n - 1). 2 + (n - 2) . 3 + ........ + 1. n

Solution

The given series is: n. 1 + (n - 1) . 2 + (n - 2) . 3 + ............. + 1. n 
Let Tk denote its kth term
∴           Tk = [kth term of n, n - 1, .....] cross times [kth term of 1, 2, 3, ........]
                 = [n + (k - 1) (-1)] [1 + (k - 1)1] = (n - k + 1) (k) = nk - k2 + k
              Tk = (n + 1)k - k2
Let space space straight S subscript straight n  denote the sum of the series

∴          straight S subscript straight n space equals space sum from straight k equals 1 to straight n of straight T subscript straight k space equals space sum from straight k equals 1 to straight n of left square bracket left parenthesis straight n plus 1 right parenthesis straight k space minus space straight k squared right square bracket space space equals space left parenthesis straight n plus 1 right parenthesis sum from straight k equals 1 to straight n of straight k space minus space sum from straight k equals 1 to straight n of straight k squared

                                        = fraction numerator left parenthesis straight n plus 1 right parenthesis straight n left parenthesis straight n plus 1 right parenthesis over denominator 2 end fraction minus 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 space equals space fraction numerator straight n left parenthesis straight n plus 1 right parenthesis over denominator 6 end fraction left square bracket 3 straight n plus 3 minus 2 straight n minus 1 right square bracket
   
                                         =   fraction numerator straight n left parenthesis straight n plus 1 right parenthesis left parenthesis straight n plus 2 right parenthesis over denominator 6 end fraction

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