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### How to get to the formula for the sum of squares of first n numbers? [duplicate]

Possible Duplicate: How do I come up with a function to count a pyramid of apples? Proof that $\sum\limits_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}$? Finite Sum of Power? I know that the sum of ...
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### How do I derive $1 + 4 + 9 + \cdots + n^2 = \frac{n (n + 1) (2n + 1)} 6$ [duplicate]

Possible Duplicate: Proof that $\sum\limits_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}$? I am introducing my daughter to calculus/integration by approximating the area under y = f(x*x) by calculating ...
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### How do I derive the formula for the sum of squares? [duplicate]

I was going over the problem of finding the number of squares in a chessboard, and got the idea that it might be the sum of squares from $1$ to $n$. Then I searched on the internet on how to calculate ...
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### Proving $\sum_{k=1}^n{k^2}=\frac{n(n+1)(2n+1)}{6}$ without induction [duplicate]

I was looking at: $$\sum_{k=1}^n{k^2}=\frac{n(n+1)(2n+1)}{6}$$ It's pretty easy proving the above using induction, but I was wondering what is the actual way of getting this equation?
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### induction proof: $\sum_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}$ [duplicate]

I encountered the following induction proof on a practice exam for calculus: $$\sum_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}$$ I have to prove this statement with induction. Can anyone please help me ...
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### how can one find the value of the expression, $(1^2+2^2+3^2+\cdots+n^2)$ [duplicate]

Possible Duplicate: Proof that $\sum\limits_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}$? Summation of natural number set with power of $m$ How to get to the formula for the sum of squares of first n ...
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### Blocks of Pyramid Pattern Expression [duplicate]

There is a pattern following, and trying to find the algebraic expression Each layer (from the top). Diagram. So the first layer has 1, second has 4, third has 9, and the fourth has 16. That's how ...
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### How do I calculate the sum of sum of triangular numbers? [duplicate]

As we know, triangular numbers are a sequence defined by $\frac{n(n+1)}{2}$. And it's first few terms are $1,3,6,10,15...$. Now I want to calculate the sum of the sum of triangular numbers. Let's ...
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### How to derive the formula for the sum of the first $n$ perfect squares? [duplicate]

How do you derive the formula for the sum of the series when $S_n = \sum_{j=1}^n j^2$? The relationship $(n, S_n)$ can be determined by a polynomial in $n$. You are supposed to use finite differences ...
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### Having trouble understanding why $\sum_{i=1}^ni^2= \frac{n(n+1)(2n+1)}{6}$ [duplicate]

So I understand $\sum\limits_{i=1}^{n}i^2 = \frac{n(n+1)(2n+1)}{6}$ but I'm not sure how to come to that conclusion. Having trouble understanding
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### Mathematical Induction: Sum of first n even perfect squares [duplicate]

So the series is $$P_k: 2^2 + 4^2 + 6^2 + ... + (2k)^2 = \frac{2k(k+1)(2k+1)}3$$ and i have to replace $P_k$ with $P_{k+1}$ to prove the series. I have to show that \frac{2k(k+1)(2k+1)}3 + [2(k+1)]...
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### The sum of the first $n$ squares $(1 + 4 + 9 + \cdots + n^2)$ is $\frac{n(n+1)(2n+1)}{6}$ [duplicate]

Prove that the sum of the first $n$ squares $(1 + 4 + 9 + \cdots + n^2)$ is $\frac{n(n+1)(2n+1)}{6}$. Can someone pls help and provide a solution for this and if possible explain the question
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I'm trying to find a way to represent $\sum_{n=1}^\infty n^2$ as a partial sum. I know that every term in this series can be represented, for example when $n=5$, as $5^2+4^2+3^2+2^2+1^2$. I know that $... 2answers 106 views ### Proof by mathematical induction$1^2+2^2+\ldots+n^2=\frac{n(n+1)(2n+1)}6$[duplicate] I am trying to solve this question. I have to proove that L = R side Question:$1^2+2^2+\ldots+n^2=\frac{n(n+1)(2n+1)}6$What I got so far, I am stuck on third step, the numbers seem too big. I do not ... 1answer 293 views ### Proving by induction that$\frac{n(n + 1)(2n + 1)}{6} = 0^2 + 1^2 + 2^2 + 3^2 + ... + n^2\$ [duplicate]

Note: I am asking this question as a simple introductory question to proofs by induction, to which I will give also my formal answer (which should be correct, if not, please comment) for future ...

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