Hello,

Somehow i dont seem to be able to solve this simple looking series...pls help..

Sum of ..... 1*(3^2) + 2(4^2) + 3(5^2) + 4(6^2) .....upto 15 terms

So, we can write the Tn as n[(n+2)^2] ,

Also
note the following about the series : as the series is 9 , 32 , 75 , 144, 245...
-- -- -- --
23 43 69 101
-- -- --
20 26 32 .....

Thanks..

2. well i think i did not post a clear post..

so as the series comes down to 9 , 32 , 75 , 144 , 245 , ...

which obviously is not an AP... but the second order differences are in AP.

32 -9 = 23
75-32 = 43 43-23= 20
144-75= 69 69-43=26
245-144=101 101-69=32 ...

Any help...??

3. Hello, skyskiers

$1\cdot3^2 + 2\cdot4^2 + 3\cdot5^2 + 4\cdot6^2 + \cdots + 15\cdot17^2$

So, we can write: . $T_k \:=\:k(k+2)^2$ . . . . yes

If you know some summation formulas, we can solve it this way . . .

$S_n \;=\;\sum^n_{k=1} k(k+2)^2$

. . $=\;\sum^n_{k=1}\left(k^3 + 4k^2 + 4k\right)$

. . $= \quad\sum^n_{k=1}k^3 \qquad + \qquad 4\sum^n_{k=1}k^2 \qquad + \qquad 4\sum^n_{k=1}k$

. . $= \;\frac{n^2(n+1)^2}{4} + 4\cdot\frac{n(n+1)(2n+1)}{6} + 4\cdot\frac{n(n+1)}{2}$

. . which simplifies to: . $S_n \;=\;\frac{n(n+1)(3n^2 + 19n + 32)}{12}$

Therefore: . $S_{15} \;=\;\frac{15\cdot16\,(3\!\cdot\!15^2 + 19\!\cdot\!15 + 32)}{12} \;=\;19,840$

4. This is how dumb can i be...

Thx for saving....Thanks a ton man!!