Question

How do you add all the odd numbers between 1-99 inclusive?

Answer 1

`sum_(n=1)^50 (2n-1)=1+3+5+7+9+......=97+99 = 2500`

Explanation:

Form a sequence `(x_n)= 1, 3, 5, 7, 9, ......, 95, 97, 99.`
This is the sequence of all the odd numbers between 1 and 99, endpoints included.

Clearly this is an arithmetic sequence with common difference d = 2 between terms.

The general term for this sequence may be given as :

`x_n=a+(n-1)d` , where a = first term, n = number of terms.

Hence, `x_n=1+(n-1)(2)`
`=2n-1`

Writing `x_n = 99` and solving for `n` gives that 99 is the 50th term.

Now to add all the terms of this sequence, yields the arithmetic series
`sum_(n=1)^50 (2n-1)=1+3+5+7+9+......=97+99.`

From the formula for sum of an arithmetic series with constant difference d, we obtain the solution as
`sum_(n=1)^n [a+(n-1)d]=n/2[2a+(n-1)d]`
`=50/2[(2)(1)+(50-1)(2)]`

`=2500`

Or, using the other formula if the first and last terms are known is:
sum`= n/2(a+l) = 50/2(1 + 99) = 2500`

Answer 2

`2500`

Explanation:

More simply, considering the sum of the series

`1 + 3 + 5 + 7 + ...+ 93 + 95 + 97 + 99`

Note that the numbers may be paired off `(1+99)`, `(3+97)`, `(5+95)`, each pair adding to `100`. There are `25` such pairs. So the sum equals `2500` `(25 xx 100)`.