Question

Frame the fornula for each of the following The sum of first n natural numbers is half the product of n and (n+1)​

Answer 1

Answer:

Shown that, the sum of first n natural numbers is $\frac{1}{2} n(n+1)$

Step-by-step explanation:

Let's assume that, the required sum = x

1 +     2 +       3 +        4 +        5 + . . . . . + (n - 4) + (n - 3) + (n - 2) + (n - 1) + n = x

n + (n - 1) + (n - 2) + (n - 3) + (n - 4) + . . . . .+  5  +      4  +       3    +      2   +  1 = x

Adding up the above two equations:

(n+1) + (n+1) + (n+1) + (n+1) + (n+1)+ . . . . .  +(n+1) + (n+1) + (n+1) + (n+1) + (n+1) = 2x

There are n number of (n+1) to be added:

Therefore, n(n+1) = 2x

Therefore, x $=\frac{1}{2} n(n+1)$