Permutations and Combinations

Permutations and Combinations

Question

Prove that the sum to n terms of the series

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Answer

Let Sn = 11 + 103 + 1005 + .......... up to n terms
          = (10 + 1) + (100 + 3) + (1000 + 5) + ............... upto n terms
          = [10 + 100 + 1000 + ..... upto n terms] + [1 + 3 + 5 + .......upto n terms]
          = [10 + 102 + 103 + ....... upto n terms] + [ 1 + 3 + 5 +........... upto n terms]
Since the series in the first bracket is a G.P. series with first term 10, common ratio 10 and the
series in the second bracket is an A.P. series with first term 1 and common difference 2.

∴     straight S subscript straight n space equals space fraction numerator 10 left parenthesis 10 to the power of straight n minus 1 right parenthesis over denominator 10 minus 1 end fraction space plus space straight n over 2 left square bracket 2.1 space plus space left parenthesis straight n minus 1 right parenthesis space 2 right square bracket space equals space 10 over 9 left parenthesis 10 to the power of straight n minus 1 right parenthesis space plus space straight n over 2 left square bracket 2 space plus space 2 straight n space minus 2 right square bracket space equals space 10 over 9 left parenthesis 10 to the power of straight n minus 1 right parenthesis space plus space straight n squared

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