Prove that the sum of the first n integers starting with 1 is equal to n*(n+1)/2
2 Answers
justaguide | College Teacher | (Level 2) Distinguished Educator
The sum of the first n integers starting with 1 is equal to `S_n = (n*(n+1))/2` . This can be proved by induction.
First, verify the formula for n = 1. `S_1 = 1` and `(1*2)/2` is also equal to 1.
Now prove that the relation is true for n + 1 if it is assumed to be true for n.
`S_(n+1) = S_n + n+1`
It has been assumed that `S_n = (n*(n+1))/2`
`S_(n+1) = S_n + n+1 `
=> `(n*(n+1))/2 + n + 1`
=> `(n*(n+1))/2 + (2*(n + 1))/2`
=> `(n+1)/2*(n+2)`
=> `((n+1)(n+2))/2`
As the formula holds for n = 1 and if it is assumed to be true for any number n it is also true for n + 1, the formula holds for all integral values of n.
The sum of the first n integers is given by the formula `S_n = (n*(n+1))/2`
aruv | High School Teacher | (Level 2) Valedictorian
Define
`2n+1=(n+1)^2-n^2`
`2n=(n+1)^2-n^2-1`
`n=(1/2){(n+1)^2-n^2-1}`
`So`
`T_n=(1/2){(n+1)^2-n^2-1}`
`` Put n=1,2,3,.......,n
`T_1=(1/2){2^2-1^2-1}`
`T_2=(1/2){3^2-2^2-1}`
`.........................`
`.........................`
`T_(n-1)=(1/2){n^2-(n-1)^2-1}`
`T_n=(1/2){(n+1)^2-n^2-1}`
`sum_(i=1)^nT_i=(1/2){(n+1)^2-1-(1+1+1+.......+1)}`
`=(1/2){(n+1)^2-1-n}`
`=(1/2)(n+1)(n+1-1)`
`=(n(n+1))/2`