A) \[{{e}^{a}}\]
B) \[\frac{{{e}^{a}}-e}{a-1}\]
C) \[(a-1){{e}^{a}}\]
D) \[(a+1){{e}^{a}}\]
Correct Answer: B
Solution :
| [b] The given series is |
| \[1+\frac{1+a}{2!}+\frac{1+a+{{a}^{2}}}{3!}+\frac{1+a+{{a}^{2}}+{{a}^{3}}}{4!}+....\] |
| Here, \[{{T}_{n}}=\frac{1+a+{{a}^{2}}+{{a}^{3}}+...to\,\,n\,\,terms}{n!}\] |
| \[=\frac{1(1-{{a}^{n}})}{(1-a)(n!)}=\frac{1}{1-a}\left( \frac{1-{{a}^{n}}}{n!} \right)\] |
| \[\therefore \,\,\,\,{{T}_{1}}+{{T}_{2}}+{{T}_{3}}+.....to\,\,\infty \] |
| \[=\frac{1}{1-a}\left[ \frac{1-a}{1!}+\frac{1-{{a}^{2}}}{2!}+\frac{1-{{a}^{3}}}{3!}+....to\,\,\infty \right]\] |
| \[=\frac{1}{1-a}\left[ \left( \frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...to\,\,\infty \right)-\left( \frac{a}{1!}+\frac{{{a}^{2}}}{2!}+\frac{{{a}^{3}}}{3!}+...to\,\,\infty \right) \right]\]\[=\frac{1}{1-a}[(e-1)-({{e}^{a}}-1)]=\frac{e-{{e}^{a}}}{1-a}=\frac{{{e}^{a}}-e}{a-1}\] |
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