Use appropriate formulae to find the sums

Question

anonymous · Answer

$$\sum_{i=1}^{50} (2i-1)^{2} $$

anonymous · Answer

I understand that there is formulas like $$\sum_{i=1}^{n} \frac{ n(n+1)(2n+1) }{ 6}$$ but how do I go about this? do I separate the (2i-1)^2

anonymous · Answer

anonymous · Answer

$$4i^{2}-4i+1$$
$$4\sum_{?}^{?}i^{2}-4\sum_{?}^{?}i+\sum_{?}^{?}1$$

anonymous · Answer

i took so long to write that. :(

anonymous · Answer

haha I creeped ya I'LL GIVE YA A MEDAL NEWAYS GURL

anonymous · Answer

i got a ridiculous answer tho.

anonymous · Answer

166650 lol wtf

anonymous · Answer

@bmelyk

anonymous · Answer

thats what i got.

anonymous · Answer

Case closed sucka

anonymous · Answer

in part b, im confused w/ the i=5.

anonymous · Answer

deja vu?

anonymous · Answer

We're in the same class..... lol btw do you know what to do if the i=5?

anonymous · Answer

not sure what you mean $i=5$

anonymous · Answer

$$\sum_{i=5}^{50}(2i-1)$$?

anonymous · Answer

okay like $$\sum_{i=5}^{20} (i^{3}-2i)$$

anonymous · Answer

its a new question lol

anonymous · Answer

Yeah, like do you ignore that and use the same formula?

anonymous · Answer

$$\sum_{i=1}^{50}(i^3-2i)-\sum_{i=1}^4(i^3-2i)$$

anonymous · Answer

$$\sum_{i=1}^{50}i^3=\left(\frac{(50)(51)}{2}\right)^2$$

anonymous · Answer

etc
compute from one to fifty, then compute from 1 to 4, subtract the second from the first

anonymous · Answer

do you  mean 1-20?

anonymous · Answer

oh it is 20. whatever
same idea

anonymous · Answer

oh okay cool thanks!

anonymous · Answer

and 1-4 to make up for the extra 4 from i=5?

anonymous · Answer

right

anonymous · Answer

check your answer by cheating http://www.wolframalpha.com/input/?i= \sum_{i%3D5}^{20}+%28i^{3}-2i%29

anonymous · Answer

hahaha evil.... thanks

anonymous · Answer

anonymous · Answer

yw