What is the summation of (i^2 + 3i - 2) with the lower index starting at i=1 and the upper index = 10 ?

Question

What is the summation of (i^2 + 3i - 2)
with lower bound limit i = 1 to upper bound n= 10?

Answer 1

∑(i = 1 to 10) (i² + 3i - 2) let it be simplified to ∑(i² + 3i - 2)

Thus the problem is
∑(i² + 3i - 2)

Using one of the rules,
∑(i² + 3i - 2) = ∑i² + ∑3i - ∑2

Simplifying,
∑(i² + 3i - 2) = ∑i² + 3∑i - 2(10)

Since i is from 1 to 10,
= 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100
+ 3(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) - 20

Adding them,
= 5 + 25 + 25 + 85 + 145 + 100 + 3(11·5) - 20
= 365 + 165
= 530

Therefore, the summation is 530.

^_^

Answer 2

summation of i^2 from 1 to x
=(1/3)x^3+(1/2)x^2+(1/6)x
summation of i from 1 to x
=(1/2)x^2+(1/2)x
summation of 1 from 1 to x
=x
summation of (i^2 + 3i - 2)
with the lower index starting at i=1
and the upper index = 10
=(1/3)10^3+(1/2)10^2+(1/6)10
+3(1/2(10^2)+1/2(10))-2*10
=385+165-20=530

Answer 3

Mathcad says 530

Answer 4

Sum (i^2) = n(n+1)(2n+1)/6 ; for i = 1...n )
Sum (i) = n(n+1)/2 ; for i = 1...n )
Sum(x) = nx ; for i = 1 ... n

whith this informatiomn you should be able to calculate your sum

anythging else you want to know ?