Linked Questions
10 questions linked to/from Sum of cubes proof
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Proving $1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$ using induction
How can I prove that
$$1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$$
for all $n \in \mathbb{N}$? I am looking for a proof using mathematical induction.
Thanks
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5
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Proof by induction: showing that two sums are equal [duplicate]
usually the tasks look like
$$\sum_{i=0}^n i^2 = \frac{(n^2+n)(2n+1)}{6}$$
or
$$\sum_{i=0}^n i^2 = i_1^2 + i_2^2 + i_3^2+...+i_n^2$$
But for the following task I have this form:
$$\left(\sum_{k=1}...
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3
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Hard Mathematical Induction [duplicate]
I have a mathematical induction question and I know what I need to do just not how to do it.
The question is:
Prove the equality of:
$$(1 + 2 + . . . + n)^2 = 1^3 + 2^3 . . . + n^3$$
Base ...
0
votes
3
answers
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proof by mathematical induction with the summation operator? [duplicate]
$$
\sum_{k=1}^n k^3 = \left( \sum_{k=1}^n k \right)^2
$$
I can't quite understand this expression, and in fact this is my biggest difficulty in finding a solution. Can someone please explain to me ?
$...
1
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0
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Why is $(1+2+3+4+...+n)^2$ equal to $1^3+2^3+3^3...+n^3$? [duplicate]
I noticed that the sum of the first $n$ cubes is equal to the square of sum of the first $n$ natural numbers:
$$
\sum\limits_{i=1}^n i^3=\frac{n^2(n+1)^2}{4}=\left(\frac{n(n+1)}{2}\right)^2=\left(\sum\...
- 751
4
votes
3
answers
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Calculating sums
My maths teacher showed me something on how to calculate sums. Let's take an example:
$$\sum_{k=1}^n k(k+1) = \sum_{k=1}^n k^2 + \sum_{k=1}^n k = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} = \frac{n(n+...
- 437
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1
answer
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Prove that $(1+2+3+\cdots+n)^2=1^3+2^3+3^3+\cdots+n^3$ $\forall n \in \mathbb{N}$. [duplicate]
Prove that $(1+2+3+\cdots+n)^2=1^3+2^3+3^3+\cdots+n^3$ for every $n \in \mathbb{N}$.
I'm trying to use induction on this one, but I'm not sure how to. The base case is clearly true. But when I add $n+...
- 41
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Proving some identities in the set of natural numbers without using induction...
I'm not sure how to prove some of the identities without using induction, for example:
$$1+2+3+...+n=\frac{n(n+1)}{2}$$
$$1^2+2^2+...+n^2=\frac{n(n+1)(2n+1)}{6}$$
$$1^3+2^3+...+n^3=(\frac{n(n-1)}{2})^...
- 1,645
0
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1
answer
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Mathematical Induction Proof 1 [duplicate]
Prove that for every integer $ n \geq 1$, we have
$\displaystyle \sum_{j=1}^n j^3 = \left(\dfrac{n(n+1)}{2}\right)^2$
I know how to prove an induction proof, but I just can't get the algebra down on ...
- 101
0
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0
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Geometrical intuition for sum of first n cubes [duplicate]
The relation
$$
\sum_{k=1}^n k^3 = \left(\sum_{k=1}^n k\right)^2
$$
baffled me when I first found out (i.e. yesterday on a train trip). Writing an inductive proof is easy and I know that there is a ...
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