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

User Calvillo
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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=\frac{81\cdot126}2\implies\boxed{S=5103}

User Siavash Rostami
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