# Prove that the sum of first n natural numbers is $$n(n+1)\over 2$$

## Solution :

Let $$S_n$$ = 1 + 2 + 3 + ….. + n = $$\sum_{k=1}^{n} k$$

Clearly, it is an arithmetic series with first term a = 1, common difference d = 1 and last term l = n.

$$\therefore$$ $$S_n$$ = $$n\over 2$$ (1 + n) = $$n(n+1)\over 2$$

### Similar Questions

Let $$a_n$$ be the nth term of an AP. If $$\sum_{r=1}^{100}$$ $$a_{2r}$$ = $$\alpha$$ and $$\sum_{r=1}^{100}$$ $$a_{2r-1}$$ = $$\beta$$, then the common difference of the AP is

A man saves Rs 200 in each of the first three months of his service. In each of the subsequent months, his saving increases by Rs 40 more than the saving of immediately previous month. His total saving from the start of service will be Rs 11040 after

If 100 times the 100th term of an AP with non-zero common difference equal to the 50 times its 50th term, then the 150th term of AP is

If x, y and z are in AP and $$tan^{-1}x$$, $$tan^{-1}y$$ and $$tan^{-1}z$$ are also in AP, then

The sum of first 20 terms of the sequence 0.7, 0.77, 0.777, ……. , is