Question

How do you find the partial sum of `Sigma (n+4)/2` from n=1 to 100?

Answer 1

`sum_(r=1)^100 (r+4)/2 = 2725`

Explanation:

We need the standard formula `sum_(r=1)^n r=1/2n(n+1)`

`:. sum_(k=1)^n (r+4)/2 = 1/2 sum_(k=1)^n (r+4)`
`" "= 1/2 {sum_(k=1)^n r +sum_(k=1)^n 4}`
`" "= 1/2 {1/2n(n+1) +4n}`
`" "= 1/2 *1/2{n(n+1) +8n}`
`" "= 1/4{n^2+n +8n}`
`" "= 1/4{n^2+9n}`
`" "= 1/4n(n+9)`

Put `n=100 =>`

`:. sum_(r=1)^100 (r+4)/2 = 1/4(100)(109)`
`" " = 2725`

Answer 2

`sum_(r=1)^100 (r+4)/2 = 2725`

Explanation:

Another approach using the AP formula:

Write out the first few terms to establish the pattern;

`sum_(r=1)^100 (r+4)/2 = 5/2 + 6/2 + 7/2 + ... 104/2`

The terms for an Arithmetic Progression (AP) with `a=5/2` and `d=1/2` and there are `100` terms.

Using the standard AP formula:

`S_n = 1/2n{2a + (n-1)d}`

We get;

`S_100 = (100)/2{2(5/2) + (99)(1/2)}`
`" " = 50{5 + 99/2}`
`" " = 2725`