Having Trouble with Summation

[The Student]

My textbook somehow has Σ (from k = 1 to n) of [(k-1)^(m+1)] equaling Σ (from k = 0 to n-1) of [k^(m+1)] which then supposedly equals 0^(m+1) = 0. I understand how it got the second expression from the first expression that I began with in this post. I also understand that if k = 0, then any power of m+1 will equal 0. But I don't understand why k seems to only equal 0. If n = 5, then depending on m the answer could be much greater than 0.

 

[tkhunny]

Are you sure we have the entire problem statement? Doesn't make much sense, so far.

 

[The Student]

Are you sure we have the entire problem statement? Doesn't make much sense, so far.

The question asks: show that for any n ∈

, Σ (from k = 1 to n) [k^(m+1) - (k - 1)^(m+1)] = n^(m+1).

Their answer, Σ (from k = 1 to n) [k^(m+1)] - Σ (from k = 1 to n) [(k - 1)^(m+1)]

= Σ (from k = 1 to n) [k^(m+1)] - Σ (from k = 0 to n - 1) [k^(m+1)]

= n^(m+1) - 0^(m+1) = n^(m+1).

 

[DrPhil]

The question asks: show that for any n ∈

, Σ (from k = 1 to n) [k^(m+1) - (k - 1)^(m+1)] = n^(m+1).

Their answer, Σ (from k = 1 to n) [k^(m+1)] - Σ (from k = 1 to n) [(k - 1)^(m+1)]

= Σ (from k = 1 to n) [k^(m+1)] - Σ (from k = 0 to n - 1) [k^(m+1)]

= n^(m+1) - 0^(m+1) = n^(m+1).

Show that \(\displaystyle \displaystyle \sum_{k=1}^n \left[k^{m+1} - (k-1)^{m+1}\right] = n^{m+1}\)

When you write the two sums separately and subtract, all but the last term of the first sum and the first term of the second sum will cancel.

Do you understand how they changed (k-1) in the second sum to k, by changing the limits of the sum?

Let j = k-1. Then \(\displaystyle \displaystyle \sum_{k=1}^n (k-1)^{m+1} = \sum_{j=0}^{n-1} j^{m+1} \)

But the index of the sum is a "dummy variable," and you can give it any name you want, such as "k". Then the two sums can be written as they showed in the second line of the proof.

 

[The Student]

Show that \(\displaystyle \displaystyle \sum_{k=1}^n \left[k^{m+1} - (k-1)^{m+1}\right] = n^{m+1}\)

When you write the two sums separately and subtract, all but the last term of the first sum and the first term of the second sum will cancel.

Do you understand how they changed (k-1) in the second sum to k, by changing the limits of the sum?

Let j = k-1. Then \(\displaystyle \displaystyle \sum_{k=1}^n (k-1)^{m+1} = \sum_{j=0}^{n-1} j^{m+1} \)

But the index of the sum is a "dummy variable," and you can give it any name you want, such as "k". Then the two sums can be written as they showed in the second line of the proof.

Oh yeah, I forgot to look at the "big picture". Thanks!!!

 

[lookagain]

The question asks: show that for any n ∈

, Σ (from k = 1 to n) [k^(m+1) - (k - 1)^(m+1)] = n^(m+1).

Their answer, Σ (from k = 1 to n) [k^(m+1)] - Σ (from k = 1 to n) [(k - 1)^(m+1)]

= Σ (from k = 1 to n) [k^(m+1)] - Σ (from k = 0 to n - 1) [k^(m+1)]

= n^(m+1) - 0^(m+1) = n^(m+1).

What I don't like about this problem is that it doesn't state what type of number m is, or what the minimum value of m is

(hence of what the minimum value of m + 1 would be). \(\displaystyle \ \ \) m must be greater than -1 for 0^(m + 1) to be defined.

While n is stated as belonging to the set of natural numbers, nothing is stated for m belonging to a particular set of numbers.

 

[The Student]

What I don't like about this problem is that it doesn't state what type of number m is, or what the minimum value of m is

(hence of what the minimum value of m + 1 would be). \(\displaystyle \ \ \) m must be greater than -1 for 0^(m + 1) to be defined.

While n is stated as belonging to the set of natural numbers, nothing is stated for m belonging to a particular set of numbers.

Welcome to my nightmare.