Question

find the sum 1.3+3.5+5.7-----------upto n term

Answer 1

given series is ....

1.3 + 3.5 + 5.7 + ...... + (2r + 1)(2r + 3)+ ....upto n terms

so, here it is clear that , $T_r=(2r-1)(2r+1)=4r^2-1$

and sum of n terms of series , $S_n=\displaystyle\Sigma^{r=n}_{r=1}{T_r}$

= $\displaystyle\Sigma^{r=n}_{r=1}{(4r^2-1)}$

= $4\displaystyle\Sigma^{r=n}_{r=1}{r^2}-\displaystyle\Sigma^{r=n}_{r=1}{1}$

we know, $4\displaystyle\Sigma^{r=n}_{r=1}{r^2}$, means sum of square of natural number = n(n + 1)(2n + 1)/6

and $4\displaystyle\Sigma^{r=n}_{r=1}{1}$ means, sum of n times of 1 i.e., 1 + 1 + 1 + 1 + 1.... n terms = n

then, Sn = 4[ n(n + 1)(2n + 1)/6 ] - n

= 2n(n + 1)(2n + 1)/3 - n

= n [ (2n + 2)(2n + 1)/3 - 1]

= n [ (4n² + 6n + 2) - 3 ]/3

= n (4n² + 6n - 1)/3

hence, sum of n terms is n (4n² + 6n - 1)/3

Answer 2

Answer:

Step-by-step explanation: