Question

How do I find the sum of the series: 4+5+6+8+9+10+12+13+14+⋯+168+169+170. since D is changing from +1, +1 to +2 ?

Answer 1

`10962`, see the explanation.

Explanation:

If we make three sums:

`4+8+12+... = sum_(k=1)^n 4k`
`5+9+13+... = sum_(k=1)^n (4k+1) = sum_(k=1)^n 4k + sum_(k=1)^n 1 = sum_(k=1)^n 4k + n`
`6+10+14+... = sum_(k=1)^n (4k+2) = sum_(k=1)^n 4k + sum_(k=1)^n 2 = sum_(k=1)^n 4k + 2n`

Then :

`sum = sum_(k=1)^n 4k + sum_(k=1)^n 4k + n + sum_(k=1)^n 4k + 2n = 3sum_(k=1)^n 4k + 3n`

`sum = 12sum_(k=1)^n k + 3n = 12 (n(n+1))/2 +3n = 6n(n+1)+3n`

`sum = 3n(2(n+1)+1) = 3n(2n+3)`

Note: sum of first `n` integers is `(n(n+1))/2`.

We have to find `n` and it's the number of members in the first sum, so:

`n=168/4 = 42`

Finally:

`sum = 3*42*(2*42+3) = 126*87 = 10962`