Linked Questions

25
votes
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 ...
8
votes
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 ...
7
votes
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?
5
votes
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 ...
1
vote
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 ...
3
votes
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 ...
1
vote
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 ...
2
votes
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 ...
0
votes
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
1
vote
3answers
726 views

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)]...
1
vote
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 $...
3
votes
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 ...
0
votes
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 ...
1
vote
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
-1
votes
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 $\forall n\in \mathbb{N}$, $$\sum ^{n}_{i=1}i^{2}=\dfrac {n\left( n+1\right) \left(2n +1\right) }{6}$$

15 30 50 per page