Question

Identify the pattern for a series and the general formula (sigma summation):

Given:

n = 1
\[-1 + xln(e)\]

n = 2
\[2 - 2xln(e) + x^{2}\ln(e)^{2}\]

n = 5
\[-120 + 120xln(e) - 60x^{2}\ln(e)^{2} + 20x^{3}\ln(e)^{3} - 5x^{4}\ln(e)^{4} + x^{5}\ln(e)^{5}\]

n = 7
\[-5040 + 5040xln(e) - 2520x^{2}\ln(e)^{2} + 840x^{3}\ln(e)^{3} - 210x^{4}\ln(e)^{4} + 42x^{5}\ln(e)^{5} - 7x^{6}\ln(e)^{6} + x^{7}\ln(e)^{7}\]

Answer 1

It gets cut off at n =7 but

\[-5040 + 5040xln(e) - 2520x^{2}\ln(e)^{2} + 840x^{3}\ln(e)^{3} - 210x^{4}\ln(e)^{4}\]
\[+ 42x^{5}\ln(e)^{5} - 7x^{6}\ln(e)^{6} + x^{7}\ln(e)^{7}\]

Answer 2

\[\Large \sum_{k=0}^{n} (-1)^{k+1} \frac{n!}{k!}x^n\ln(e)^k\]

Answer 3

(-1)^(0+1) is -1...if n = 2, then you have -2x^2 * 1

Answer 4

\[\Huge \sum _{k=0} ^{n} (-1)^{n-k} \frac{n!}{k!} x^k \ln(e)^k\]