For each postive integer $n:$ $a_n=\frac{n^2}{n^2-45n+675}.$ Evaluate $a_1+2a_2+3a_3+\cdots+44a_{44}$

Question

For each postive integer $n:$ $a_n=\frac{n^2}{n^2-45n+675}.$ Evaluate $a_1+2a_2+3a_3+\cdots+44a_{44}$

What I have tried: I have taken $a_1,a_2,\cdots,a_{44}$ and put these values into $a_1+2a_2+3a_3+\cdots+44a_{44}.$

Is there any other way to solve this problem?

Any guidance will be highly appreciated.Thank you!

Answer 1

Based on what Hari Shankar mentioned above:

Try pairing $ka_k+(n−k)a_{n-k}$.

Here $n = 45$.
Doing that will yield: $$\sum_{k=1}^{22}\left(\frac{k^3}{k^2-45k+675} + \frac{(45-k)^3}{(45-k)^2 -45(45-k) + 675}\right) = \sum_{k=1}^{22}135=2970.$$

Answer 2

$a_{45-n}=\frac{(45-n)^2}{(45-n)^2-45(45-n)+675}= \frac{(45-n)^2}{n^2-45n+675}$

The denominator is the same! So $na_n+(45-n)a_{45-n}=\frac{n^3}{n^2-45n+675}+ \frac{(45-n)^3}{n^2-45n+675}$. Now we use the fact that $x^3+y^3=(x+y)(x^2-xy+y^2)$:$$ \begin{align} na_n+(45-n)a_{45-n}&=\frac{(n+(45-n))(n^2-n(45-n)+(45-n)^2)}{n^2-45n+675}\\ &=\frac{45(n^2-45n+n^2+45^2-90n+n^2)}{n^2-45n+675}\\ &=\frac{45(3n^2-135n+45^2)}{n^2-45n+675}\\ &=135 \end{align} $$ So each pair works out to a constant value, $135$ and there are $22$ pairs total. The sum is $2970$