Question

Sum the series: $199\cdot 1+197\cdot 3+195\cdot 5+.....+103\cdot 197+101\cdot 99$.

Answer 1

$199.1+197.3+195.5+...+1.99$

$a_n=(-2n+201)(2n-1)$

$a_n=-4n^2+2n+402n-201$

$a_n=4n^2+404n-201$

$\sum a_n=\sum (-4n^2+404m)-201$

$\sum a_n=-4\sum n^2+404\sum n-\sum 201$

$\sum a_n=(-4)\frac{n(n+1)(2n+1)}{6}+\frac{404n(n+1)}{2}-201n$

$n=50$

$\sum _{n=1}^{50}{a}_n(-4)\frac{50(50+1)\left(101\right)}{6}+404\frac{50(50+1)}{2}-201\times 50$

$\sum _{n=1}^{50}{a}_n=(-4)25\times 17\times 101+404\times 25\times 51-201\times 50$

$=-1,71,700+5,15,100-10,050$

$=-1,81,750+5,15,100$

$=3,33,350$

$\therefore Ans3,33,350$