Find the sum to n terms of the | Class 11 Mathematics Chapter Sequence and Series, Sequence and Series NCERT Solutions

Question:

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

Answer:

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

nth term, an = n ( n + 1) ( n + 2)

= (n2 + n) (n + 2)

n3 + 3n2 + 2n

   S subscript n space equals space sum from k equals 1 to n of space a subscript k
space space space space space space
space space space space space space equals space sum from k equals 1 to n of space k cubed space plus space 3 space sum from k equals 1 to n of space k squared space plus space 2 sum from k equals 1 to n of space k
space space space space space space equals space open square brackets fraction numerator n open parentheses n plus 1 close parentheses over denominator 2 end fraction close square brackets squared space plus space fraction numerator 3 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 2 n open parentheses n plus 1 close parentheses over denominator 2 end fraction
space space space space space space equals space open square brackets fraction numerator n open parentheses n plus 1 close parentheses over denominator 2 end fraction close square brackets squared space space plus space fraction numerator n open parentheses n plus 1 close parentheses open parentheses 2 n plus 1 close parentheses over denominator 2 end fraction space plus space n open parentheses n plus 1 close parentheses
space space space space space space equals space fraction numerator n open parentheses n plus 1 close parentheses over denominator 2 end fraction open square brackets fraction numerator n open parentheses n plus 1 close parentheses over denominator 2 end fraction space plus space 2 n space plus 1 plus 2 close square brackets
space space space space space space equals space fraction numerator n open parentheses n plus 1 close parentheses over denominator 2 end fraction open square brackets fraction numerator n squared space plus n plus 4 n plus 6 over denominator 2 end fraction close square brackets
space space space space space space space equals space fraction numerator n open parentheses n plus 1 close parentheses over denominator 4 end fraction open parentheses n squared space plus 5 n space plus 6 close parentheses
space space space space space space space space equals space fraction numerator n open parentheses n plus 1 close parentheses over denominator 4 end fraction open parentheses n squared space plus space 2 n space plus space 3 n space plus 6 close parentheses
space space space space space space space space equals fraction numerator n open parentheses n plus 1 close parentheses open square brackets n open parentheses n plus 2 close parentheses plus 3 open parentheses n plus 2 close parentheses close square brackets over denominator 4 end fraction
space space space space space space space space space equals fraction numerator n open parentheses n plus 1 close parentheses open parentheses n plus 2 close parentheses open parentheses n plus 3 close parentheses over denominator 4 end fraction


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  • Read the question carefully and focus on the core concept being asked.
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Welcome to the NCERT Solutions for Class 11 Mathematics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 4 , Question 2: Find the sum to n terms of the series 1 × 2 × 3 + 2 × 3 × 4 + 3 &t....