If sum_{r=1}^{25} left( frac{r}{r^{4}+r^{2}+1 | vedclass

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(A) We have the term $T_r = \frac{r}{r^4+r^2+1}$.Using the identity $r^4+r^2+1 = (r^2+r+1)(r^2-r+1)$,we can write:$T_r = \frac{r}{(r^2+r+1)(r^2-r+1)}

Vedclass · Accepted Answer

$976$

Vedclass · Answer

(A) We have the term $T_r = \frac{r}{r^4+r^2+1}$.Using the identity $r^4+r^2+1 = (r^2+r+1)(r^2-r+1)$,we can write:$T_r = \frac{r}{(r^2+r+1)(r^2-r+1)} = \frac{1}{2} \left( \frac{1}{r^2-r+1} - \frac{1}{r^2+r+1} \right)$.Let $f(r) = \frac{1}{r^2-r+1}$. Then $f(r+1) = \frac{1}{(r+1)^2-(r+1)+1} = \frac{1}{r^2+2r+1-r-1+1} = \frac{1}{r^2+r+1}$.Thus,$T_r = \frac{1}{2} (f(r) - f(r+1))$.The sum is $S = \sum_{r=1}^{25} T_r = \frac{1}{2} \sum_{r=1}^{25} (f(r) - f(r+1))$.This is a telescoping sum: $S = \frac{1}{2} (f(1) - f(26))$.$f(1) = \frac{1}{1^2-1+1} = 1$.$f(26) = \frac{1}{26^2-26+1} = \frac{1}{676-26+1} = \frac{1}{651}$.$S = \frac{1}{2} (1 - \frac{1}{651}) = \frac{1}{2} (\frac{650}{651}) = \frac{325}{651}$.Since $\gcd(325, 651) = 1$,we have $p=325$ and $q=651$.Therefore,$p+q = 325 + 651 = 976$.

If $\sum_{r=1}^{25} \left( \frac{r}{r^{4}+r^{2}+1} \right) = \frac{p}{q}$ where $p$ and $q$ are positive integers such that $\gcd(p,q)=1$,then $p+q$ is equal to . . . . . . .

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