Question

If $\sum _{r=1}^{r=n}\frac{r^4+r^2+1}{r^4+r}=\frac{675}{26},$ then $n$ is equal to

• A. $10$
• B. $15$
• C. $25$
• D. $30$

Answer 1

Answer: (C)

$25$

$t_r=\frac{(r^4+r)+(r^2-r+1)}{(r^4+r)}=1+\frac{r^2-r+1}{r(r+1)(r^2-r+1)}=1+\frac{1}{r(r+1)}=1+(\frac{1}{r}-\frac{1}{r+1})$
Thus, the terms are:
$r=1:1+(\frac{1}{1}-\frac{1}{2})$
$r=2:1+(\frac{1}{2}-\frac{1}{3})$
$r=3:1+(\frac{1}{3}-\frac{1}{4})$
...
$r=n:1+(\frac{1}{n}-\frac{1}{n+1})$
Hence, the sum: $S=n+(\frac{1}{1}-\frac{1}{n+1})=\frac{n(n+2)}{n+1}$
Thus: $\frac{n(n+2)}{n+1}=\frac{675}{26}=>n=25$
Hence, (c) is correct.