Question

Use​ Gauss's approach to find the following sums​ a. 1+2+3+4..+999 b. 1+3+5..+997

Answer 1

`S=1+2+3+4+\cdots+996+997+998+999`

`S=999+998+997+996+\cdots+4+3+2+1`

`\implies 2S=(1+999)+(2+998)+\cdots+(998+2)+(999+1)`

Note that we're adding 999 terms in
`S`, so
`2S` is summing together 999 pairs of numbers that add to 1000:


`2S=1000+1000+\cdots+1000+1000`

`2S=999(1000)`

`S=\frac{999(1000)}2=999(500)=499500`


`T=1+3+5+\cdots+993+995+997`

`T=997+995+993+\cdots+5+3+1`

`\implies 2T=(1+997)+(3+995)+\cdots+(995+3)+(997+1)`

This time,
`T` adds up 499 numbers - we can determine this by finding the value of
`n` such that
`2n-1=997` - that each add up to 998, so


`2T=998+998+\cdots+998+998`

`2T=499(998)`

`T=\frac{499(998)}2=499(499)=249001`