Consider the telescoping series given below:

${\displaystyle \underset{n=1}{\overset{}{}}}{c}_n={\displaystyle \underset{n=1}{\overset{}{}}}(\frac{7}{(n+3{)}^3}-\frac{7}{(n+1{)}^3})$

Write the expressions for the first four terms $in$ the sequence $of$ partial sums $\{s_m\}.$ You may cancel

positive and negative versions $of$ the same fraction, but otherwise, leave your answers $as$ unsimplified sums

$of$ fractions. $Do$ not try $to$ combin$\frac{e}{s}$implify fractions $(other$ than the cancelation mentioned earlier$)or$ use

decimal approximations.

$s_1=$

$s_2=$

$s_3=\sqrt[\phantom{2}]{(\frac{7}{(1+3{)}^3}-\frac{7}{(1+1{)}^3})+(\frac{7}{(2+3{)}^3}-\frac{7}{(2+1{)}^3})+(\frac{7}{(3+3{)}^3}-\frac{7}{(3+1{)}^3})}$

$s_4=$

$(\frac{7}{(1+3{)}^3}-\frac{7}{(1+1{)}^3})+(\frac{7}{(2+3{)}^3}-\frac{7}{(2+1{)}^3})+(\frac{7}{(3+3{)}^3}-\frac{7}{(3+1{)}^3})+(\frac{7}{(4+3{)}^3}-\frac{7}{(4+1{)}^3})$

$s_5=$

$(\frac{7}{(1+3{)}^3}-\frac{7}{(1+1{)}^3})+(\frac{7}{(2+3{)}^3}-\frac{7}{(2+1{)}^3})+(\frac{7}{(3+3{)}^3}-\frac{7}{(3+1{)}^3})+(\frac{7}{(4+3{)}^3}-\frac{7}{(4+1{)}^3})+(\frac{7}{(5+3{)}^3}-\frac{7}{(5+1{)}^3})$

Based $on$ the data above, give a formula for the $m^{th}$ partial sum, $s_{m}m2$