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  • Find the sum to n terms of the series in Exercises 8 to 10 whose nth terms is given by n (n+1) (n+4).

8.   Find the sum to n terms of the series in Exercises 8 to 10 whose nth terms is given by n (n+1) ( n + 4 )

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nth terms is given by n (n+1) ( n + 4 )

a_n=n (n+1) ( n + 4 )=n(n^2+5n+4)=n^3+5n^2+4n

S_n=\sum _{k=1}^{n} a_k=\sum _{k=1}^{n} k(k+1)(k+4)

                        =\sum _{k=1}^{n} k^3+5\sum _{k=1}^{n} k^2+4\sum _{k=1}^{n} k

                     =\left [ \frac{n(n+1)}{2} \right ]^2+\frac{5.n(n+1)(2n+1)}{6}+\frac{4.n(n+1)}{2}

                   =\left [ \frac{n(n+1)}{2} \right ]^2+\frac{5.n(n+1)(2n+1)}{6}+2.n(n+1)

                  =\left [ \frac{n(n+1)}{2} \right ] (\frac{n(n+1)}{2}+\frac{5(2n+1)}{3}+4)

                 =\left [ \frac{n(n+1)}{2} \right ] \left ( \frac{3n^2+3n+20n+10+24}{6} \right )

                =\left [ \frac{n(n+1)}{2} \right ] \left ( \frac{3n^2+23n+34}{6} \right )

                =\left [ \frac{n(n+1)}{24} \right ] \left ( 3n^2+23n+34 \right )

Posted by

seema garhwal

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