# Linked Questions

2answers
130k views

### 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 ...
5answers
14k views

### 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 ...
5answers
2k views

### 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?
3answers
8k views

### 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 ...
3answers
3k views

### 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 ...
4answers
1k views

### 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 ...
1answer
3k views

### 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 ...
2answers
243 views

### 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 ...
2answers
400 views

### 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
3answers
726 views

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 $$\frac{2k(k+1)(2k+1)}3 + [2(k+1)]... 4answers 77 views ### How to represent this partial sum? [duplicate] I'm trying to find a way to represent \sum_{n=1}^\infty n^2 as a partial sum. I know that every term in this series can be represented, for example when n=5, as 5^2+4^2+3^2+2^2+1^2. I know that ... 1answer 227 views ### Proving by induction that \frac{n(n + 1)(2n + 1)}{6} = 0^2 + 1^2 + 2^2 + 3^2 + … + n^2 [duplicate] Note: I am asking this question as a simple introductory question to proofs by induction, to which I will give also my formal answer (which should be correct, if not, please comment) for future ... 2answers 94 views ### Finding an expression for the sum of n tems of the series 1^2 + 2^2 + 3^2 + … + n^2 [duplicate] Possible Duplicate: why is \sum\limits_{k=1}^{n} k^m a polynomial with degree m+1 in n I know that if you have a non-arithmetic or geometric progression, you can find a sum S of a series ... 3answers 120 views ### The sum of the first n squares (1 + 4 + 9 + \cdots + n^2) is \frac{n(n+1)(2n+1)}{6} [duplicate] Prove that the sum of the first n squares (1 + 4 + 9 + \cdots + n^2) is \frac{n(n+1)(2n+1)}{6}. Can someone pls help and provide a solution for this and if possible explain the question 1answer 68 views ### How to prove using math induction that \forall n\in \mathbb{N}, \sum ^{n}_{i=1}i^{2}=\frac{1}{6}n\left( n+1\right) \left(2n +1\right)? [duplicate] Use mathematical induction to prove that \forall n\in \mathbb{N},$$\sum ^{n}_{i=1}i^{2}=\dfrac {n\left( n+1\right) \left(2n +1\right) }{6}

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