If _r=1^25(rr^4+r^2+1)=(p)/(q), where p and q are positive integers such that gcd(p,q)=1, …

Question

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

Answer: 976

Accepted Answer

Given

$\sum _{r=1}^{25}(\frac{r}{{r}^{4}+{r}^{2}+1})=\frac{p}{q}$

The expression ${r}^{4}+{r}^{2}+1$ can be factored using a completion of squares technique:

${r}^{4}+{r}^{2}+1=({r}^{4}+2{r}^{2}+1)-{r}^{2}={({r}^{2}+1)}^{2}-{r}^{2}$

Using the difference of squares formula, ${a}^{2}-{b}^{2}=(a-b)(a+b)$ :

${r}^{4}+{r}^{2}+1=({r}^{2}-r+1)({r}^{2}+r+1)$

Split into Partial Fractions

Now, let's rewrite the general term ${T}_{r}$ :

${T}_{r}=\frac{r}{({r}^{2}-r+1)({r}^{2}+r+1)}$

We can express the numerator $r$ in terms of the factors in the denominator:

$({r}^{2}+r+1)-({r}^{2}-r+1)=2r$

So, $r=\frac{1}{2}[({r}^{2}+r+1)-({r}^{2}-r+1)]$. Substituting this back:

${T}_{r}=\frac{1}{2}[\frac{({r}^{2}+r+1)-({r}^{2}-r+1)}{({r}^{2}-r+1)({r}^{2}+r+1)}]$

${T}_{r}=\frac{1}{2}[\frac{1}{{r}^{2}-r+1}-\frac{1}{{r}^{2}+r+1}]$

Evaluate the Telescoping Sum

Let $f(r)=\frac{1}{{r}^{2}-r+1}$.

Note that $f(r+1)=\frac{1}{(r+1{)}^{2}-(r+1)+1}=\frac{1}{{r}^{2}+2r+1-r-1+1}=\frac{1}{{r}^{2}+r+1}$.

Our sum becomes : $S=\frac{1}{2}\sum _{r=1}^{25}[f(r)-f(r+1)]$

Expanding this:

For $r=1$ : $f(1)-f(2)$

For $r=2:f(2)-f(3)$

For $r=3:f(3)-f(4)$

\ldots

For $r=24:f(24)-f(25)$

For $r=25:f(25)-f(26)$

Result of summation: $\sum _{r=1}^{25}[f(r)-f(r+1)]=f(1)-f(26)$

So,

$S=\frac{1}{2}[f(1)-f(26)]$

Calculate the Final Value

$f(1)=\frac{1}{{1}^{2}-1+1}=1$

$\begin{gathered} f(26)=\frac{1}{{25}^{2}+25+1}=\frac{1}{625+25+1}=\frac{1}{651} \ S=\frac{1}{2}[1-\frac{1}{651}]=\frac{1}{2}(\frac{651-1}{651})=\frac{650}{2\times 651} \ S=\frac{325}{651} \end{gathered}$

$325=(5{)}^{2}\times 13$ and $651=3\times 7\times 31$ they share no common factor $gcd(325,651)=1$

So $p=325$ and $q=651$$\therefore $$p+q=325+651=976$

Solutions · 1