Question

Here is a small challenge for your entertainment and gratification. For any natural number n, let f(n) denote the sum of the numbers from 1 to n. Thus f(1)=1, f(2)=1+2=3, f(3)=1+2+3=6, f(100)=1+2+3+…+100=5050, etc.
It turns out that f is a polynomial of degree 2 in n. Figure out the coefficients of f:
f(n)=​

Answer 1

$\textbf{Given:}$

$\textsf{f(1)=1, f(2)=3, . . .  . .}$

$\textsf{and f(n) is the sum of numbers from 1 to n}$

$\textbf{To find:}$

$\textsf{f(n)}$

$\textbf{Solution:}$

$\textsf{Consider,}$

$\mathsf{f(n)}$

$\mathsf{=Sum\;of\;numbers\;from\;1\;to\;n}$

$\mathsf{=1+2+3+\;.\;.\;.\;.+n}$

$\mathsf{=\dfrac{n(n+1)}{2}}$

$\mathsf{=\dfrac{n^2+n}{2}}$

$\implies\boxed{\mathsf{f(n)=\dfrac{n^2+n}{2}}}$