\sum_{r=1}^{n}r^3=\frac{n^2}{4}\right\left(n+1)\right^2
show that
\sum_{r=n+1}^{2n}r^3=\frac{n^2}{4}\right\left(3n+1)\right\left(5n+3)\right
I don't understand how to start
It isn't homework, it's in a textbook and I'm having trouble with it.
When r=1, summing to n the series of r^3 = (n^2)/4 (n+1)^2
Show that when r = (n+1), summing to 2n = (n^2)/4 (3n+1)(5n+3)
What order do you start the summation, and how do I begin?
If the fourth, seventh and sixteenth terms of an AP are in geometric progression, the first six terms of the AP have a sum of 12, find the common difference of the AP and the common ratio of the GP.
I've been assuming that the fourth, seventh and sixteenth terms of the AP are the fourth...
This isn't a homework question, it's in a textbook I have and I'm a bit stumped. I know there's something relatively simple I'm missing so any help would be much appreciated (working too).
Three consecutive terms of an A.P. have a sum of 36 and a product of 1428. Find the three terms.