Question

Prove that the sum of the first n integers starting with 1 is equal to n*(n+1)/2

Accepted Answer

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

Answer

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

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Prove that the sum of the first n integers starting with 1 is equal to n*(n+1)/2

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