# Help with math Progressions

1. May 8, 2005

(1) Lets say you have the progression $5 + 12 + 21 + ... + 1048675$ and you want to find the sum of 20 terms. I know that the kth term is given by $2^{k} +3 + 5(k-1)$. So would I treat the $2^{k}$ terms separately from the $3 + 5(k-1)$ terms? Would it be $2 + 4 + 8 + 16 +... + 2^{n}$ and $3 + 8 + 13 + ... + (3+5(k-1))$. Would the total sum be $\frac{2 - 2(2)^{n}}{-1} + \frac{n}{2}(3+ (3+5(k-1))$?

$$2097248 + 1010$$

(2) $$3 + 10 + 25 + ... + 39394$$ and you want to find sum of first 10 terms. I know that the kth term is $2 \times 3^{k-1} + 1 + 3(k-1)$ Would i do the same thing and treat the $3^{k-1}$ and $1 + 3(k-1)$ separately?

Thanks

Last edited: May 8, 2005
2. May 8, 2005

### The Bob

I have only looked at (1) but here is what I have done:

a1 = 5
a2 = 12
a3 = 21

You have said that: ak = 2k + 5k - 2

So: ak + 1 = 2(k+1) + 5(k+1) - 2

This means that: ak+1 + ak = 2(k+1) + 5(k+1) - 2 + 2k + 5k - 2 = 2k + 1 + 2k + 10k + 1

and: This means that: ak+2 + ak+1 + ak = 2(k+2) + 5(k+2) - 2 + 2(k+1) + 5(k+1) - 2 + 2k + 5k - 2
= 2(k+2) + 2(k+1) + 2k + 15k + 9

I have found that the sum of the terms 2(k+n) for n = 0, 1, 2 etc. is 2(k+1) - 2.

The term for the sum of the 5k, 10k, 15k terms is equal to 5k2

However I was unable to work out the last term. All you need to do, from what I have done, is work out the way to express the difference between -2, 1 and 9 in terms of k when k is 1, 2 and 3 respectively. Then you can put all three terms together and you can find the sum up to any kth term.