Question 1

Find the sum to n terms of the series 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …

Answer

The given series is 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …

nth term, an = n ( n + 1)

∴ S subscript n space equals space sum from k equals 1 to n of a subscript k space equals space sum from k equals 1 to n of k open parentheses k plus 1 close parentheses

space space space space space equals space sum from k equals 1 to n of space k squared space plus space sum from k equals 1 to n of space k
space space space space space equals space fraction numerator n open parentheses n plus 1 close parentheses open parentheses 2 n plus 1 close parentheses over denominator 6 end fraction space plus space fraction numerator n open parentheses n plus 1 close parentheses over denominator 2 end fraction
space space space space space equals fraction numerator n open parentheses n plus 1 close parentheses over denominator 2 end fraction open parentheses fraction numerator 2 n plus 1 over denominator 3 end fraction space plus space 1 close parentheses
space space space space space equals space fraction numerator n open parentheses n plus 1 close parentheses over denominator 2 end fraction open parentheses fraction numerator 2 n plus 4 over denominator 3 end fraction close parentheses
space space space space space equals space fraction numerator n open parentheses n plus 1 close parentheses open parentheses n plus 2 close parentheses over denominator 3 end fraction

 

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