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

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##