I was studying sequence and series and used the formula many times $$1+2+3+\cdots +n=\frac{n(n+1)}{2}$$ I want its proof.
Thanks for any help.
Answer
Let the sum be $$S_n=1+2+3+\cdots +n\tag1$$ on reversing the same equation we get $$S_n=n+(n-1)+(n-2)+\cdots +1\tag2$$ On adding $(1)$ and $(2)$ we have each term equal to $n+1$ which will occur $n$ times i.e. $$2S_n=(n+1)+(n+1)+(n+1)\cdots\{n times\}+(n+1)$$ $$2S_n=n(n+1)$$ $\therefore$ $$S_n=\frac{n(n+1)}{2}.$$ Hope it helps!!!
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