Math, asked by SunilKumar3625, 1 month ago

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)=​

Answers

Answered by Anonymous
1

{\frak{\underline{Given:-}}}

</p><p>\tt{f(1)=1, f(2)=3, .  .  . .}f(1)=1,

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

{\frak{\underline{To \:  find::-}}}

\texttt{f(n)}

{\frak{\underline{Solution:-}}}

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

\mathtt{=1+2+3+\;.\;.\;.\;.+n}

\mathtt{=\dfrac{n(n+1)}{2}}=

\mathtt{=\dfrac{n^2+n}{2}}=

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

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