prove that 1+2+3+.........+n=(n+1)÷2
Anonymous:
what assumption?
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1+2+3+...+n is an Arithmetic progression with common difference of 1.
so we can find the sum by formula....
Sn =n/2[a+l]...
where l is the last term of the AP.
so summation of first n natural numbers...
Sn=n/2[1+n]=n(n+1)/2
so we can find the sum by formula....
Sn =n/2[a+l]...
where l is the last term of the AP.
so summation of first n natural numbers...
Sn=n/2[1+n]=n(n+1)/2
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