Question

16. Find the sum of the first n positive integers.​

Answer 1

1+n = n+1

2+(n-1) = n+1

3+(n-2) = n+1

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(n-1)+2 = n+1

n+1 = n+1

on the left we are adding one number that scans 1 through n to another number that scans the same range backwards. on the right we always have (n+1)

adding up all these equations we get the sum of the number 1 through n twice on the left and n(n+1) on the right. so,

1+2+3 +....+n = n(n+1)/2

Answer 2

Answer:

sn = n/2 (2a + (n-1)d )

hope it helps