Find the value of the sum.
n integral i=1
(3+4i)^2

Question

Find the value of the sum.
n integral i=1 
(3+4i)^2

anonymous · Answer

summoning @sirm3d  :D

sirm3d · Answer

can you type the using the equation button?

sirm3d · Answer

or maybe draw

anonymous · Answer

Ok one second.

anonymous · Answer

They really should make an app for iPad...

anonymous · Answer

$$\sum_{i=1}^{n}(3+4i)^2$$

sirm3d · Answer

we'll need three formulas
$$\sum_{i=1}^{n}k=kn$$ $$\sum_{i=1}^{n}i=\frac{ n(n+1) }{ 2 }$$ $$\sum_{i=1}^{n}i^2 =\frac{ n(n+1)(2n+1) }{ 6 }$$

sirm3d · Answer

just expand the squared binomial, then apply the formula above

anonymous · Answer

9+ 24i + 16 i^2 and then sub i and i^2 into those formulas?

sirm3d · Answer

yes

anonymous · Answer

attachment

anonymous · Answer

i'm no done yet, but is it suppose to be like tis?

sirm3d · Answer

it should be (n+1), not (n-1)

anonymous · Answer

oh..memorized it wrong.. ok..doing it again

sirm3d · Answer

just change the minus sign in your computation

anonymous · Answer

yea, i did. is everything else correct? just need to combine like terms?

sirm3d · Answer

yep. and don't miss the term n^3

anonymous · Answer

16/3 n^3 + 20n^2 + 29n :)

sirm3d · Answer

got it right

anonymous · Answer

thanks :DD

anonymous · Answer

anonymous · Answer

the answer i got is 16/3 n^3 + 20n^2 + 71/3 n

sirm3d · Answer

integral is for continuous variables, summation is for discrete variables

anonymous · Answer

oohh. ok got it :) ty

anonymous · Answer

then there is no way to check my answer :/

sirm3d · Answer

here's a fourth formula $$\sum_{i=1}^{n}i^3=n^2(n+1)^2/4$$

anonymous · Answer

but there is no i^3 in my function

sirm3d · Answer

is this a summation problem or an integral problem?

sirm3d · Answer

the link gave the equation 4x^3-2x+2