Question #33e60

Suppose #1+2+3+...+55=S#
then
#color(white)(2" ")S=color(white)(" ")1+color(white)("")2+""3+...+53+54+55#
#color(white)(2" ")ul(S)=ul(55+54+53+...+3+" "2+1)#
#2S=underbrace(56+56+56+...+56+56+56)_(55" terms")#

#2S=55 xx 56#

#S=(55 xx 56)/2=1540#

Here is a common way to find the sum of consecutive numbers:
#1+2+3+…+55#

Group it like this way:
#=(1+55)+(2+54)+(3+53)+…+(27+29)+28#

Note that, if there are an even number of numbers, then each number will get paired with another one. If, as in this case, there are an odd number of numbers, one number will be left out.

So,
#=56+56+56+…+56+28#
and there are #27# of the #56#'s (since originally there were #55# numbers, and except for one, the remaining #54# were paired; thus, there are #54/2=27# pairs).

#=27*56+28#
#=1540#

Let's try to find a more general way to find the sum of consecutive numbers. Notice that when we paired the numbers off, the sum of each pair was equal. Since the first pair is the sum of the smallest and largest number, all of the pairs had a sum equal to the smallest number in the sequence added by the largest number in the sequence.

Now, there are two possible cases. The first case is when the number of numbers is even. Then, all of the numbers are paired together, and the number of pairs is the total number of numbers divided by #2#. So, since each pair is equal to the smallest number in the sequence added by the largest number in the sequence, the sum of the sequence is half of the total number of numbers multiplied by the sum of the smallest number in the sequence and the largest number in the sequence.

But what if the number of numbers is odd? Realize that the number of pairs is still half of the number of numbers. Even though the number of numbers is not divisible by #2#, when the numbers are paired, the odd number left out is exactly half of the sum of a pair (for #1+2+3+…+55#, the sum of each pair is #56#, and the number left out is #28#). So, the sum of the sequence is half of the total number of numbers multiplied by the sum of the smallest number in the sequence and the largest number in the sequence, exactly the same for the case when the number of numbers is even.

To summarize, when you have a sequence of consecutive numbers, the sum is half of the total number of numbers, multiplied by the sum of the smallest number in the sequence and the largest number in the sequence.

So, #1+2+3+…+55#. The sum is equal to half of the total number of numbers, #55/2#, multiplied by the sum of the smallest number in the sequence and the largest number of the sequence #1+55#. Thus, the sum is #55/2*(1+55)=1540#, same as what we got before.