Expand the series and Evaluate:
Enter your answer as the example: 1+2+3+4+5=15
I got 225 as an answer but I need the numbers that add up to it
you know to evaluate it, yes? just state the terms it adds up to. e.g., evaluate it for k=1. then k=2. k=3 so on to k=5. then, add them all together but don't simplify to the final product
No I have no idea how to evaluate it
n^2(n+1)^2 /4
so for n=5, 25*36 / 4 =225
What?? I have no idea what I'm doing
theres a formula for sums of cubes until the nth term. n^2(n+1)^2 /4 use it with k=5 or develop like : 1^3+2^3+3^3+4^3+5^3
I don't understand what do I do with k=5?
Holy pellet please help me
just plug it in the formula and you get the answer. or expand the series as I said
WHAT AM I SUPPOSED TO PLUG IN THIS LOOKS LIKE GIBBERISH
flutter
I got 1+9+36+100+225 but its wrong
\( \displaystyle \sum_{k=1}^{5} k^3 \) Expanding this out just means we evaluate the summand k^3 for each k value between 1 and 5, and add them all up. For k = 1, k^3 = 1^3 = 1 For k =2, k^3 = 2^3 = 8 <-- this is cube and not square. For k =3, k^3 = 3^3. Calculate this as well. Doing the same process for k=4, and 5. The sum expanded out is thus: \( \displaystyle \sum_{k=1}^{5} k^3 = 1^3 + 2^3 + 3^3 + \cdots \)
The result is what that expanded form evaluates into after adding the numbers, which does come out to 225.
Ok I know the answer is 225 I need the numbers that it adds up to
You know k^3 means k * k * k ? Like, 3^3 = 3 * 3 * 3 = 27 ?
So its 1^2+2^2+3^3+4^2+5^5? ITS WRONG
1^3+2^3+3^3+4^3+5^3=225 is wrong
It is the same exponent for all of them. 1^3, 2^3, 3^3, 4^3, and 5^3. This is from k^3. \( \displaystyle \sum_{k=1}^{5} k^3 = \underbrace{1^3}_{k=1} + \underbrace{2^3}_{k=2} + \underbrace{3^3}_{k=3} + \underbrace{4^3}_{k=4} + \underbrace{5^3}_{k=5} \) But did you evaluate each one for the actual number for each? Not 1^3, but 1. Not 2^3, but 8. etc.
I got 1+8+27+64+125=225 and its right thank you for being patient with me
No worries. Glad to help. :)
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