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123. Find the sum of n terms of the series whose nth term is

$space space fraction numerator 1 cubed plus 2 cubed plus 3 cubed plus...... plus straight n cubed over denominator straight n end fraction.$

Here, $straight T subscript straight n space equals space fraction numerator 1 cubed plus 2 cubed plus.... plus straight n cubed over denominator straight n end fraction space equals space fraction numerator sum straight n cubed over denominator straight n end fraction space equals space fraction numerator straight n squared left parenthesis straight n plus 1 right parenthesis squared over denominator 4 straight n end fraction space equals space fraction numerator straight n left parenthesis straight n plus 1 right parenthesis squared over denominator 4 end fraction space equals space straight n over 4 left parenthesis straight n squared plus 2 straight n plus 1 right parenthesis$

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∴ $straight S subscript straight n space equals space sum from straight k equals 1 to straight n of straight T subscript straight k space equals space sum from straight k equals 1 to straight n of open parentheses 1 fourth straight k cubed plus 1 half straight k squared plus 1 fourth straight k close parentheses space equals space 1 fourth sum from straight k equals 1 to straight n of straight k cubed plus 1 half sum from straight k equals 1 to straight n of straight k squared space plus space 1 fourth sum from straight k equals 1 to straight n of straight k$

 $equals 1 fourth fraction numerator straight n squared left parenthesis straight n plus 1 right parenthesis squared over denominator 4 end fraction plus 1 half fraction numerator straight n left parenthesis straight n plus 1 right parenthesis left parenthesis 2 straight n plus 1 right parenthesis over denominator 6 end fraction plus 1 fourth fraction numerator straight n left parenthesis straight n plus 1 right parenthesis over denominator 2 end fraction equals fraction numerator straight n left parenthesis straight n plus 1 right parenthesis over denominator 4 end fraction open square brackets fraction numerator straight n left parenthesis straight n plus 1 right parenthesis over denominator 4 end fraction plus fraction numerator 2 straight n plus 1 over denominator 3 end fraction plus 1 half close square brackets$

$equals space fraction numerator straight n left parenthesis straight n plus 1 right parenthesis over denominator 4 end fraction open square brackets fraction numerator 3 straight n squared plus 3 straight n plus 8 straight n plus 4 plus 6 over denominator 12 end fraction close square brackets space equals space fraction numerator straight n left parenthesis straight n plus 1 right parenthesis over denominator 48 end fraction space left square bracket 3 straight n squared plus 11 straight n plus 10 right square bracket$

$space space equals space fraction numerator straight n left parenthesis straight n plus 1 right parenthesis over denominator 48 end fraction space left square bracket 3 straight n squared plus 6 straight n plus 5 straight n plus 10 space equals space fraction numerator straight n left parenthesis straight n plus 1 right parenthesis over denominator 48 end fraction space left square bracket 3 straight n left parenthesis straight n plus 2 right parenthesis space plus space 5 left parenthesis straight n plus 2 right parenthesis right square bracket$

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128 Views

Long Answer Type

124.

Find the sum to n terms of the series:

1 . 2 . 4 + 2. 3 . 7 + 3. 4. 10 +..............

161 Views

Short Answer Type

125. Find the sum of the cubes of first n odd natural numbers. Or Sum the series to n terms l³ + 3³ + 5³ +..........

90 Views

126.

Find the sum to n terms the series $fraction numerator 1 over denominator 1.2 end fraction plus fraction numerator 1 over denominator 2.3 end fraction plus fraction numerator 1 over denominator 3.4 end fraction plus...........$ and deduce the sum to infinity.

102 Views

127.

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

94 Views

Long Answer Type

128. Show that the sum of the cubes of any number of consecutive integers is divisible by the sum of these integers.

154 Views

Short Answer Type

129.

Show that $fraction numerator 1 cross times 2 squared plus 2 cross times 3 squared plus...... plus straight n cross times left parenthesis straight n plus 1 right parenthesis squared over denominator 1 squared cross times 2 plus 2 squared cross times 3 plus...... plus straight n squared cross times left parenthesis straight n plus 1 right parenthesis end fraction space equals space fraction numerator 3 straight n plus 5 over denominator 3 straight n plus 1 end fraction$

157 Views

130.

Find the sum to n terms of the following series:

5² + 6² + 7² + ....+ 20².

137 Views