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

2 Answers

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

Jan 16, 2017

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).