Question on writing summations in expanded form

[mr_coffee]

Hello everyone.

I have the following:

THe sumnation of 1/k! from k = 0, to n. I"m suppose to write this in expanded form.

I did the following:
1/0! + 1/1! + 1/2! + 1/3! +...+1/n! = 1 + 1 + 1/2 + 1/6 + 1/24 + 1/120 +...+ 1/n!

Is that what they wanted?
An example simliar to this one was the following:
Sumnation (-2)^i from i = 1 to n.
(-2)^1 + (-2)^2 + (-2)^3 + ... + (-2)^n = -2 + 2^2 -2^3 +...+(-1)^n(2)^n

from mine the signs don't seem to be changing, so is it just simply 1/n!

Also i was wondering if someone could check to see if i did this one correctly:
Write each using summnation or product notation.

41. n + (n-1)/2! + (n-2)/3! + (n-3)/4! + .. + 1/n!

I said:
Sumnation from k = 1 to n, (n-k)/n!Thanks!

 

[HallsofIvy]

Hello everyone.

I have the following:

THe sumnation of 1/k! from k = 0, to n. I"m suppose to write this in expanded form.

I did the following:
1/0! + 1/1! + 1/2! + 1/3! +...+1/n! = 1 + 1 + 1/2 + 1/6 + 1/24 + 1/120 +...+ 1/n!

Is that what they wanted?

Yes, that is correct.

An example simliar to this one was the following:
Sumnation (-2)^i from i = 1 to n.
(-2)^1 + (-2)^2 + (-2)^3 + ... + (-2)^n = -2 + 2^2 -2^3 +...+(-1)^n(2)^n

from mine the signs don't seem to be changing, so is it just simply 1/n!

Also i was wondering if someone could check to see if i did this one correctly:
Write each using summnation or product notation.

41. n + (n-1)/2! + (n-2)/3! + (n-3)/4! + .. + 1/n!

I said:
Sumnation from k = 1 to n, (n-k)/n!


Thanks!

Not quite. n is a fixed number, k is changing so the denominator is not n!, it is k!.

 

[mr_coffee]

Thanks for the help!