Question

how do we find f(x), if given the summation of f(x) is equal to 2n^2+3n.. hint: summation n=(1/2)(n)(n+1)

Answer 1

find f(x) if\[\sum_{1}^{n}f(x)= 2n ^{2}+3n\], given\[\sum_{1}^{n}r = \frac{ 1 }{ 2 }n(n+1)\]

Answer 2

\[\sum_{1}^{n}r =\frac{ n }{ 2 }(n+1)\]
Multiplying by 4 on both sides
\[\sum_{1}^{n}4r = 2n(n+1) = 2n^{2} + 2n\]
Adding n on both sides:
\[n + \sum_{1}^{n}4r = 2n ^{2} + 3n\]
\[or, \sum_{1}^{n}(4r+1) = 2n^{2} + 3n\]

Answer 3

so f(x) = 4x+1