Question

Calculus help?

Answer 1

@phi , @ranga @surjithayer ?

Answer 2

Are you supposed to prove the summation is 2970?

Answer 3

I am supposed to solve it and I'm guessing it is 2970? but how do I solve it?

Answer 4

Are you given the formula for: i = 1 to n of sigma (i^3) ?

Answer 5

there are some theorems

Answer 6

i'll attach them

Answer 7

∑▒〖i(i^2-1)=∑▒i^3 - ∑▒i=(i^(2 ) (i+1)^2)/4〗-i(i+1)/2
Put i=10 and get the solution
[ ∑▒n^3 =(∑▒n)^2
∑▒n=n(n+1)/2 ]

Answer 8

The problem asks you to find the sum of:
i = 1 to 10 of Sigma i * (i^2 - 1)
Since we will make use of the theorems, let us do a general summation by using "n" in the place of 10:
Find:
{ i = 1 to n of Sigma i * (i^2 - 1) } =
{ i = 1 to n of Sigma i^3 } - { i = 1 to n of Sigma i }
substitute the formula from the theorems:
= n^2(n+1)^2/4 - n(n+1)/2
Put n = 10
= 10^2 * (10 + 1)^2 / 4 - 10(10+1)/2 =
100 * 121 / 4 - 10 * 11 / 2 =
3025 - 55 = 2970

Answer 9

Thank you so much. I get confused as to which theorem to use.

Answer 10

Thanks again!

Answer 11

You are welcome.