sum of the series from n =1 to ∞ whose nth term is 1/(n+1)! a) e b) e-2 c)e+1 d) e-3
in Algebra 1 Answers by

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
To avoid this verification in future, please log in or register.

1 Answer

 We know the derivative of eˣ is eˣ.

If we write eˣ=a₀+a₁x+a₂x²+...+a_nx^n

When x=0, 1=a₀.

Now differentiate:

eˣ=a₁+2a₂x+...+na_nx^(n-1)

and put x=0:

1=a₁.

Differentiate again:

eˣ=2a₂+...n(n-1)x^(n-2).

When x=0, 1=2a₂.

If we repeatedly differentiate, and put x=0 to solve for a_n, we end up with the series:

1+x+x²/2!+x³/3!+...+x^n/n! to infinity.

Put x=1 and e=1+1+1/2!+1/3!+...+1/n!.

e-2=1/2!+1/3!+...

So the general term is 1/(n+1)!, the required series.

Therefore (b) is the answer.

 

 

by Top Rated User (652k points)

Related questions

Welcome to MathHomeworkAnswers.org, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
83,040 questions
87,759 answers
1,974 comments
4,590 users