The sum of the following series 

1+6+\frac{9(1^{2}+2^{2}+3^{2})}{7}+\frac{12(1^{2}+2^{2}+3^{2}+4^{2})}{9}+\frac{15(1^{2}+2^{2}+3^{2}+.....+5^{2}}{11})+.......

up to 15 terms, is:

  • Option 1)

     

    7830

  • Option 2)

     

    7820

  • Option 3)

     

    7510

  • Option 4)

     

    7520

Answers (1)
A admin

 

Summation of series of natural numbers -

\sum_{k=1}^{n}K= \frac{1}{2}n\left ( n+1 \right )
 

- wherein

Sum of first n natural numbers

1+2+3+4+------+n= \frac{n(n+1)}{2}

 

 

Summation of series of natural numbers -

\sum_{k=1}^{n}K^{2}= \frac{1}{6}n\left ( n+1 \right )\left ( 2n+1 \right )
 

- wherein

Sum of  squares of first n natural numbers

1^{2}+2^{2}+3^{2}+4^{2}+------+n^{2}= \frac{n(n+1)\left ( 2n+1 \right )}{6}

 

 

Summation of series of natural numbers -

\sum_{k=1}^{n}K^{3}=\left ( \sum_{k=1}^{n}K \right )^{2}

- wherein

1^{3}+2^{3}+3^{3}+-------n^{3}\\= \left ( 1+2+3+-----+n \right )^{2}

 

The general term of the sequence will be

T_{n} = \frac{(3+(n-1)\times 3)(1^2 +2^2 +3^2+...+n^2)}{(2n+1)}

T_{n} = \frac{3\cdot \frac{n(n+1)(2n+1)}{6}}{(2n+1)} = \frac{n^2(n+1)}{2}

S_15 = \sum_{n=1}^{15}T_n = \frac{1}{2}\sum_{n=1}^{15}(n^3 + n^2) \\ = \frac{1}{2}\left[\left(\frac{15(15+1)}{2} \right )^2 + \frac{15\times 16\times 31}{6} \right ] = 7820


Option 1)

 

7830

Option 2)

 

7820

Option 3)

 

7510

Option 4)

 

7520

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