Question

Find $\frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+...16$ terms.

• A. $123$
• B. $\frac{1785}{4}$
• C. $\frac{1875}{4}$
• D. $234$

Answer 1

Answer: (A), (B)

$\frac{1785}{4}$

Since, $1^3+2^3+.......+n^3=\frac{n^2(n+1{)}^2}{4}$ .....(i) and
$1+3+5+.......+(2n-1)=n^2$......(ii)
Thus, $T_n=\frac{1^3+2^3+.......+n^3}{1+3+5+.......+(2n-1)}=\frac{n^2(n+1{)}^2}{4n^2}=\frac{1}{4}(n+1{)}^2$
As we know that $\sum n^2=\frac{n(n+1)(2n+1)}{2}$........(iii)
Replace $n$ by $(n+1)$,
Hence sum is $=\sum T_n=\frac{1}{4}\sum (n+1{)}^2$

$=\frac{1}{4}\cdot \frac{(n+1)(n+2)\left(2\right(n+1)+1)}{6}$

$=\frac{(n+1)(n+2)(2n+3)}{24}$
For required sum put $n=16$, we get the required sum,

$=\frac{17\left(18\right)\left(35\right)}{24}$
$=\frac{17\times 3\times 35}{4}$
$=\frac{1785}{4}$