Discrete Math - finding the sum of a sequence

[shannonw05]

I have to find the sum of the sequence

(i^2 + 4i - 5) from i=1 to i=20.

I figured out that the difference is equal to

the previous term + (2n-3)

I went online and saw that the formula for the sum of an arithmetic sequence is

.5n(first term - nth term)

however when I use this formula I don't get the right answer. Hopefully this makes sense. Any advice would be wonderful! Thanks

 

[galactus]

Use the formulae for the sums of squares, sums of successive integers.

\(\displaystyle \L\\\sum_{i=1}^{20}{i^{2}}=\frac{i(2i+1)(i+1)}{6}=\frac{20(2\cdot20+1)(20+1)}{6}\)

\(\displaystyle \L\\\sum_{i=1}^{20}{i}=\frac{i(i+1)}{2}=\frac{20(20+1)}{2}\)

\(\displaystyle \L\\\sum_{i=1}^{20}5=5i=5(20)\)

 

[shannonw05]

So am I supposed to add up the answer to each formula to find the sum? Because it's 4i do I need to multiply each "i" in the second formula by 4? I was under the impression that there was one formula that would yield the answer. Whenever I try adding the answers to the three formulas I'm still not getting the correct answer. [/quote]

 

[galactus]

Yes, you multiply the 2nd part by 4.

Add them up. I'll use n instead of i:

\(\displaystyle \L\\\frac{n(n+1)(2n+1)}{6}+4\cdot\frac{n(n+1)}{2}-5n=\frac{2n^{3}+15n^{2}-17n}{6}\)

Sub in n=20 and you find the answer is 3610

 

[shannonw05]

Thanks so much! You've been a huge help.