How Do You Calculate the Sum of Specific Mathematical Progressions?

[courtrigrad]

(1) Let's 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

 

[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: a_k = 2^k + 5k - 2

So: a_{k + 1} = 2^(k+1) + 5(k+1) - 2

This means that: a_k+1 + a_k = 2^(k+1) + 5(k+1) - 2 + 2^k + 5k - 2 = 2^{k + 1} + 2^k + 10k + 1

and: This means that: a_k+2 + a_k+1 + a_k = 2^(k+2) + 5(k+2) - 2 + 2^(k+1) + 5(k+1) - 2 + 2^k + 5k - 2
= 2^(k+2) + 2^(k+1) + 2^k + 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 k^th term.

The Bob (2004 ©)

 

[thepatientmental]

for reaching out for help with math progressions! To find the sum of the first 20 terms in the first progression, you can use the formula for the sum of a finite geometric series: S_n = a(1-r^n)/(1-r), where a is the first term and r is the common ratio. In this case, a=5 and r=2. So the sum would be S_20 = 5(1-2^20)/(1-2) = 5(1-1048576)/(-1) = 5(1048575) = 5242875.

For the second progression, you can use the same formula to find the sum of the first 10 terms. Here, a=3 and r=3. So the sum would be S_10 = 3(1-3^10)/(1-3) = 3(1-59049)/(-2) = 3(59048)/(-2) = -88572.