Question

Helppppppp
3
Σ (2n – 3)
n=0

Answer 1

Since there are only 4 terms in the sum, it's not too much work to expand it as

`\displaystyle \sum_(n=0)^3 (2n-3) = -3 + (-1) + 1 + 3 = \boxed{0}`

Alternatively, we can use the well-known formulas

`\displaystyle \sum_(n=1)^N 1 = \underbrace{1 + 1 + \cdots + 1}_{N \text{ 1s}} = N`

`\displaystyle \sum_(n=1)^N n = 1+2+\cdots+N = \frac{N(N+1)}2`

These sums start at n = 0, so in our given sum we will keep track of the 0-th term separately:

`\displaystyle \sum_(n=0)^3 (2n-3) = -3 + \sum_(n=1)^3 (2n-3) = -3 + 2 \sum_(n=1)^3 n - 3 \sum_(n=1)^3 1 = -3 + 3*4 - 3*3 = 0`

as expected.