Question

find the sum of 23+24+25+...+103

Answer 1

Let

$S=23+24+25+\cdots+101+102+103$

This sum has ___ terms. Its terms form an arithmetic progression starting at 23 with common difference between terms of 1, so that the $n$-th term is given by the sequence $23+(n-1)\cdot1=22+n$. The last term is 103, so there are

$103=22+n\implies n=81$

terms in the sequence.

Now, we also have

$S=103+102+101+\cdots+25+24+23$

so that adding these two ordered sums together gives

$2S=(23+103)+(24+102)+\cdots+(102+24)+(103+23)$

$\implies2S=\underbrace{126+126+\cdots+126+126}_{81\text{ times}}=81\cdot126$

$\implies S=\dfrac{81\cdot126}2\implies\boxed{S=5103}$