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`1+(2x)/(1!)+(3x^(2))/(2!)+(4x^(3))/(3!)+..infty` is equal toA. `xe^(x)`B. `(x+1)e^(x)`C. `xe^(-x)`D. `(x+1)e^(-x)` |
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Answer» Answer: We have `1+(2x)/(1!)+(3x^(2))/(2!)+(4x^(3))/(3!)+..infty` `=underset(n=0)overset(infty)Sigma (n+1)x^(n)/(n!)=underset(n=0)overset(infty)Sigma(n)/(n!)x^(n)+underset(n=0)overset(infty)Sigma(x^(n))/(n!)` `=xunderset(n=1)overset(infty)Sigma (x^(n-1))/(n-1)!+ underset(n=0)overset(infty)Sigma (x^(n))/(n!)=xe^(x)+e^(x)=(x+1)e^(x)` |
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