Exponential Theorem – Problems
Example 1: Show that \(1+\frac{1+5}{2!}+\frac{1+5+{{5}^{2}}}{3!}+…=\frac{1}{(4)}\left( {{e}^{5}}-e \right)\).
Solution: Given that,
\(1+\frac{1+5}{2!}+\frac{1+5+{{5}^{2}}}{3!}+…=\frac{1}{(4)}\left( {{e}^{5}}-e \right)\).
L.H.S: \(1+\frac{1+5}{2!}+\frac{1+5+{{5}^{2}}}{3!}+…\).
\(1+\frac{1+5}{2!}+\frac{1+5+{{5}^{2}}}{3!}+…=\sum\limits_{n=1}^{\infty }{\frac{1+5+{{5}^{2}}+…+{{5}^{n-1}}}{n!}}\).
\(=\sum\limits_{n=1}^{\infty }{\frac{{{5}^{n}}-1}{(5-1)n!}}\).
\(=\sum\limits_{n=1}^{\infty }{\frac{{{5}^{n}}-1}{(4)n!}}\).
\(=\frac{1}{(4)}\sum\limits_{n=1}^{\infty }{\frac{{{5}^{n}}-1}{n!}}\).
\(=\frac{1}{(4)}\sum\limits_{n=1}^{\infty }{\frac{{{5}^{n}}}{n!}}-\frac{1}{(4)}\sum\limits_{n=1}^{\infty }{\frac{1}{n!}}\).
\(=\frac{1}{(4)}\left( {{e}^{5}}-1 \right)-\frac{1}{(4)}\left( e-1 \right)\).
\(=\frac{1}{(4)}\left( {{e}^{5}}-e \right)\).
Example 2: Show that \(1+\frac{1+a}{2!}+\frac{1+a+{{a}^{2}}}{3!}+…=\frac{{{e}^{a}}-e}{a-1}\).
Solution: Given that,
\(1+\frac{1+a}{2!}+\frac{1+a+{{a}^{2}}}{3!}+…=\frac{{{e}^{a}}-e}{a-1}\).
L.H.S: \(1+\frac{1+a}{2!}+\frac{1+a+{{a}^{2}}}{3!}+…\).
\(1+\frac{1+a}{2!}+\frac{1+a+{{a}^{2}}}{3!}+…=\sum\limits_{n=1}^{\infty }{\frac{1+a+{{a}^{2}}+…+{{a}^{n-1}}}{n!}}\).
\(\sum\limits_{n=1}^{\infty }{\frac{1+a+{{a}^{2}}+…+{{a}^{n-1}}}{n!}}=\sum\limits_{n=1}^{\infty }{\frac{{{a}^{n}}-1}{(a-1)n!}}\).
\(=\frac{1}{(a-1)}\sum\limits_{n=1}^{\infty }{\frac{{{a}^{n}}-1}{n!}}\).
\(=\frac{1}{(a-1)}\sum\limits_{n=1}^{\infty }{\frac{{{a}^{n}}}{n!}}-\frac{1}{(a-1)}\sum\limits_{n=1}^{\infty }{\frac{1}{n!}}\).
\(=\frac{1}{\left( a-1 \right)}\left[ \left( {{e}^{a}}-1 \right)-\left( e-1 \right) \right]=\frac{\left( {{e}^{a}}-e \right)}{\left( a-1 \right)}\).
Hence proved \(1+\frac{1+a}{2!}+\frac{1+a+{{a}^{2}}}{3!}+…=\frac{{{e}^{a}}-e}{a-1}\).