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  • Find the sum to n terms of each of the series 3 × 8 + 6 × 11 + 9 × 14 + ...

6.  Find the sum to n terms of each of the series 3 \times 8 + 6 \times 11 + 9\times 14+...

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series = 3 \times 8 + 6 \times 11 + 9\times 14+...

           =(n th term of 3,6,9,...........)\times(nth terms of 8,11,14,..........)  

n th term  = 3n(3n+5)=a_n=9n^2+15n

S_n=\sum _{k=1}^{n} a_k=\sum _{k=1}^{n} 3k(3k+5)

                        =9\sum _{k=1}^{n} k^2+15\sum _{k=1}^{n} k

                     =\frac{9.n(n+1)(2n+1)}{6}+\frac{15.n(n+1)}{2}

                     =\frac{3.n(n+1)(2n+1)}{2}+\frac{15.n(n+1)}{2}  

                   =\frac{n(n+1)}{2}\left (3(2n+1)+15 \right )

                   =\frac{3.n(n+1)}{2}\left (2n+1+5 \right )

                  =\frac{3.n(n+1)}{2}\left (2n+6\right )

                =\frac{3.n(n+1)}{2}.2.\left (n+3\right )

                 =3.n(n+1)\left (n+3\right )

Hence, sum is  =3.n(n+1)\left (n+3\right )

Posted by

seema garhwal

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