An algebra problem by Aditya Moger

$\begin{aligned} a & = \frac {1^2}1 + \frac {2^2}3+ \frac {3^2}5 + ... + \frac {1001^2}{2001} \\ b & = \frac {1^2}3 + \frac {2^2}5+ \frac {3^2}7 + ... + \frac {1001^2}{2003} \\ \implies a - b & = \frac {1^2}1 + \frac {2^2-1^2}3+ \frac {3^2-2^2}5 + ... + \frac {1001^2-1000^2}{2001} - \frac {1001^2}{2003} \\ & = 1 + \frac {(2+1)(2-1)}3+ \frac {(3+2)(3-2)}5 + ... + \frac {(1001+1000)(1001-1000)}{2001} - \frac {1001^2}{2003} \\ & = 1 + \frac {(\cancel 3)(1)}{\cancel 3}+ \frac {(\cancel 5)(1)}{\cancel 5} + ... + \frac {(\cancel {2001})(1)}{\cancel {2001}} - \frac {1001^2}{2003} \\ & = \underbrace{1+1+1+...+1}_{1001 \times 1} - \frac {1001^2}{2003} \\ & = 1001 - \frac {1001^2}{2003} \\ & \approx \boxed{501} \end{aligned}$

$\displaystyle a-b = 1 +\left[\sum_{i=1}^{1000}\dfrac{(i+1)^2-i^2}{2i+1}\right]-\dfrac{1001^2}{2003}$

$\displaystyle \phantom{a-b} = 1 +\left[\sum_{i=1}^{1000}\dfrac{i^2+2i+1-i^2}{2i+1}\right]-\dfrac{1001^2}{2003}$

$\displaystyle \phantom{a-b} = 1 +\left[\sum_{i=1}^{1000}\dfrac{2i+1}{2i+1}\right]-\dfrac{1001^2}{2003}$

$\displaystyle \phantom{a-b} = 1 +\left[\sum_{i=1}^{1000}1\right]-\dfrac{1001^2}{2003}$

$\displaystyle \phantom{a-b} = 1 +1000 -\dfrac{1001^2}{2003}$

$\displaystyle \phantom{a-b} \approx \boxed{501}$

Substituting $n=1$ into $a_{m+n}=a_m+a_n+mn$ :

$a_{m+1}=a_m+a_{1}+m.$

Since $a_1 = 1$ , $a_{m+1}=a_m+m+1$ . Therefore, $a_m = a_{m-1} + m, a_{m-1}=a_{m-2}+(m-1), a_{m-2} = a_{m-3} + (m-2)$ , and so on until $a_2 = a_1 + 2$ .

Adding the Left Hand Sides of all of these equations gives $a_m + a_{m-1} + a_{m-2} + a_{m-3} + \cdots + a_2$ ; adding the Right Hand Sides of these equations gives $(a_{m-1} + a_{m-2} + a_{m-3} + \cdots + a_1) + (m + (m-1) + (m-2) + \cdots + 2)$ .

These two expressions must be equal; hence $a_m + a_{m-1} + a_{m-2} + a_{m-3} + \cdots + a_2 = (a_{m-1} + a_{m-2} + a_{m-3} + \cdots + a_1) + (m + (m-1) + (m-2) + \cdots + 2)\\ a_m = a_1 + (m + (m-1) + (m-2) + \cdots + 2).$

Substituting $a_1 = 1$ :

$a_m = 1 + (m + (m-1) + (m-2) + \cdots + 2) = 1+2+3+4+ \cdots +m = \frac{(m+1)(m)}{2}$ . Thus we have a general formula for $a_m$ and substituting $m=12$ : $a_{12} = \frac{(13)(12)}{2} = (13)(6) = 501$ .