finite series

• Feb 18th 2009, 12:33 AM
thereddevils
finite series
If S=1(n)+(2n-1)+3(n-2)+...+r(n+1-r)+...+n(1)

and T=1(n-1)+2(n-2)+3(n-3)+...+r(n-r)+...+(n-1)(1) , where n is a positive integer , prove that

S+T=1/6n(n+1)(2n+1)

I can see the pattern , where :

T=1(n-1+1)+2(n-2+1)+3(n-3+1) and so on but it doesn't lead me anywhere further .

Thanks .
• Feb 18th 2009, 12:26 PM
Moo
Hello,
Quote:

Originally Posted by thereddevils
If S=1(n)+(2n-1)+3(n-2)+...+r(n+1-r)+...+n(1)

and T=1(n-1)+2(n-2)+3(n-3)+...+r(n-r)+...+(n-1)(1) , where n is a positive integer , prove that

S+T=1/6n(n+1)(2n+1)

I can see the pattern , where :

T=1(n-1+1)+2(n-2+1)+3(n-3+1) and so on but it doesn't lead me anywhere further .

Thanks .

$S=\sum_{r=1}^n r(n+1-r)$
$T=\sum_{r=1}^{n-1} r(n-r)$

$S+T=\sum_{r=1}^n r(n+1-r)+\sum_{r=1}^{n-1} r(n-r)$
now make the sums begin and finish at the same values :
$S+T=n+\sum_{r=1}^{\color{red}n-1} r(n+1-r)+\sum_{r=1}^{n-1} r(n-r)$

$S+T=n+\sum_{r=1}^{n-1} [r(n+1-r)+r(n-r)]$
$S+T=n+\sum_{r=1}^{n-1} r(n+1-r+n-r)$
$S+T=n+\sum_{r=1}^{n-1} r(2n+1-2r)$
$S+T=n+(2n+1) \sum_{r=1}^{n-1} r-2 \sum_{r=1}^{n-1} r^2$
$S+T=n+\frac 12 (2n+1)(n-1)n-2 \cdot \frac{(n-1)n(2n-1)}{6} \quad {\color{red}\star}$
$S+T=n+\frac 16 \cdot n(n-1) \cdot [3(2n+1)-2(2n-1)]$
$S+T=n+\frac 16 \cdot n(n-1)(2n+5)$
$S+T=\frac n6 \cdot (6+(n-1)(2n+5))$
$S+T=\frac n6 \cdot (2n^2+3n+1)$
$S+T=\frac{n(2n+1)(n+1)}{6}$

${\color{red}\star}$ is got by these formulae :
$\sum_{j=1}^k j=\frac{k(k+1)}{2}$
and $\sum_{j=1}^k j^2=\frac{k(k+1)(2k+1)}{6}$
here, k=n-1

Of course, all of these could be solved by induction :D
• Feb 18th 2009, 01:27 PM
Jester
Maybe a little easier

$S=\sum_{r=1}^n r(n+1-r)$
$T=\sum_{r=1}^{n-1} r(n-r) =\sum_{r=1}^{n} r(n-r)$

So following Moo
$T+ S = \sum_{r=1}^n r(n+1-r) + r(n-r) = \sum_{r=1}^n r(2n+1)-2r^2$ $= \frac{n(n+1)(2n+1)}{2} - 2\frac{n(n+1)(2n+1)}{6}$ giving the answer.