Calculate the sum from k=1 to infinity (or n) of
(k^4)- (k-1)^4
It says that the solution is n^4, but I do not know why, I tried separating it in a telescoping sum, and expanding it, but I obtain at the end more values than expected. Please help
look at it. Each term adds a new k^4 term, but then subtracts out the previous term ( k-1)^4. So only the last term is left....which will be n^4
Sorry! The real problem was:
(k^4)- (k-4)^4
Yeah I understood what you said, does it apply the same here?
To calculate the sum from k=1 to infinity of (k^4) - (k-1)^4, we can simplify the expression and then look for any patterns.
Let's start by expanding (k-1)^4 using the binomial theorem:
(k-1)^4 = (k-1)(k-1)(k-1)(k-1)
= (k-1)(k^3 - 4k^2 + 6k - 4)
= k^4 - 4k^3 + 6k^2 - 4k - k^3 + 4k^2 - 6k + 4
Notice that there are several terms that cancel out when we subtract this from k^4:
-4k^3 and +k^3 cancel each other out, leaving -3k^3,
6k^2 and +4k^2 cancel each other out, leaving 10k^2,
-4k and -6k cancel each other out, leaving -10k,
and finally, +4 cancels out.
Hence, the expression (k^4) - (k-1)^4 simplifies to -3k^3 + 10k^2 - 10k.
Now, let's look at the sum from k=1 to n:
𝑆 = (1^4 - 0^4) + (2^4 - 1^4) + (3^4 - 2^4) + ... + (n^4 - (n-1)^4)
If we expand this sum, we get:
𝑆 = 1^4 - 0^4 + 2^4 - 1^4 + 3^4 - 2^4 + ... + n^4 - (n-1)^4
Grouping the terms, we notice that most of them cancel out:
𝑆 = 1^4 - 0^4 + (2^4 - 1^4) + (3^4 - 2^4) + ... + (n^4 - (n-1)^4)
We can see that all the terms except the first and last ones cancel out:
𝑆 = 1^4 - 0^4 + n^4
Since 1^4 - 0^4 equals 1, we have:
𝑆 = 1 + n^4
Therefore, the sum from k=1 to infinity (or n) of (k^4) - (k-1)^4 is n^4.