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

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May 27, 2016

Answer:

#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#

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Write your answer here...
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Answer

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Explanation

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26
Jan 16, 2017

Answer:

#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)#.

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