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

##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##