Question

How do you find the sum of the arithmetic series `Sigma(11+2k)` from k=1 to 150?

Answer 1

`S_150 = sum_(k=1)^150(11+2k) = 24300`

Explanation:

Method 1 - Using the AP Series

The sum of the first `n` terms of an AP whose terms are

`a, a+d, a+2d, ... , a+(n-1)d`

is given by

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

So for `sum_(k=1)^150(11+2k)` the terms are 13, 15, 17, ... , 311

So the terms for an AP with `n=150, a=13,d=2`

`:. S_150 = 150/2{2(13)+(149)(2)}`
`:. S_150 = 75(26+298)`
`:. S_150 = 75(324)`
`:. S_150 = 24300`

Method 2 - Using the standard results:

`sum_(r=1)^n r = 1/2n(n+1)`

We have;

`sum_(k=1)^150(11+2k) = sum_(k=1)^150(11) + 2sum_(k=1)^150(k)`
`:. sum_(k=1)^150(11+2k) = (150)(11) + 2 1/2(150)(151)`
`:. sum_(k=1)^150(11+2k) = 1650 + 22650`
`:. sum_(k=1)^150(11+2k) = 24300`, as before

Method 3 - Derive a General Formula

`sum_(k=1)^n(11+2k) = sum_(k=1)^n(11) + 2sum_(k=1)^n(k)`
`sum_(k=1)^n(11+2k) = 11n + 2 1/2n(n+1)`
`sum_(k=1)^n(11+2k) = 11n + n(n+1)`
`sum_(k=1)^n(11+2k) = n(11 + n + 1)`
`sum_(k=1)^n(11+2k) = n(n+12)`

And when `n=150 => sum_(k=1)^150(11+2k) = (150)(162) = 24300`, as before