1. ## Sum The Series

How would I solve this?:

Sum The Series-

1 x 2 + 2 x 3 + 3 x 4 + . . . + 1994 x 1995

Thanx

2. $\displaystyle \displaystyle\sum_{k=1}^{1994}k(k+1)=\sum_{k=1}^{1 994}(k^2+k)=\sum_{k=1}^{1994}k^2+\sum_{k=1}^{1994} k$
Now use the formulas
$\displaystyle \displaystyle\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6 }$ and $\displaystyle \displaystyle\sum_{k=1}^nk=\frac{n(n+1)}{2}$

3. Hello, red_dog!

Great solution!

$\displaystyle \displaystyle\sum_{k=1}^{1994}k(k+1) \;= \;\sum_{k=1}^{1994}(k^2+k)\;=\;\sum_{k=1}^{1994}k^ 2+\sum_{k=1}^{1994}k$

Now use the formulas:

$\displaystyle \displaystyle\sum_{k=1}^nk^2\;=\;\frac{n(n+1)(2n+1 )}{6}$ .and .$\displaystyle \displaystyle\sum_{k=1}^nk\;=\;\frac{n(n+1)}{2}$
I took it a step further . . .

$\displaystyle \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \;=\; \frac{n(n+1)(2n + 1 + 3)}{6} \;=\;\frac{n(n+1)(n+2)}{3}$

A curious result: the product of three consecutive integers divided by 3.

4. This result can be generalized to n consecutive sums of products.
(Likewise, with reciprocals too).