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How can I show that these are equal? [duplicate]

Sum of First $n$ Squares Equals $frac{n(n+1)(2n+1)}{6}$Prove that $sum_{k=1}^{n}{k^3}= left(frac{n(n+1)}{2}right)^2$Next step to show that these matrice expressions are equal?Prove summations are equalHow are these equations equal?Can somone help me do this double sum problem. I know how to do it manually, but I would like to know how to do it using summation formulas.show that $q$ and $r$ are unique when $r$ is less than or equal to zero.Why are these sums equal?are these summations equalHow can I show that the Lucas numbers are given by the sum of $F_{k-1}+F_{k+1}$Show that these two sums are equalHow are these 2 sums equal?

0

$begingroup$

This question already has an answer here:

• 
  Sum of First $n$ Squares Equals $frac{n(n+1)(2n+1)}{6}$
  
  31 answers
  
  

How can I show that:$$sum_{k=1}^{n}({k^2})$$
Is equal to: $$frac{n(n+1)(2n+1)}{6}$$
I know that I would apply the sum formula, should I also be using this formula? $$sum_{k=1}^{n}k=frac{n(n+1)}2$$

discrete-mathematics summation

share|cite|improve this question

edited Mar 11 at 19:45

gt6989b

34.9k22557

asked Mar 11 at 19:32

Usama GhawjiUsama Ghawji

666

$endgroup$

marked as duplicate by Dietrich Burde, mfl, gt6989b, Community♦ Mar 11 at 19:47

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  "I know that I would apply the sum formula" What is "the sum formula". "should I also be using this formula?" Isn't that formula the sum formula?
  $endgroup$
  – fleablood
  Mar 11 at 19:47
  

  
  
  
  

add a comment | 

0

$begingroup$

This question already has an answer here:

• 
  Sum of First $n$ Squares Equals $frac{n(n+1)(2n+1)}{6}$
  
  31 answers
  
  

How can I show that:$$sum_{k=1}^{n}({k^2})$$
Is equal to: $$frac{n(n+1)(2n+1)}{6}$$
I know that I would apply the sum formula, should I also be using this formula? $$sum_{k=1}^{n}k=frac{n(n+1)}2$$

discrete-mathematics summation

share|cite|improve this question

edited Mar 11 at 19:45

gt6989b

34.9k22557

asked Mar 11 at 19:32

Usama GhawjiUsama Ghawji

666

$endgroup$

marked as duplicate by Dietrich Burde, mfl, gt6989b, Community♦ Mar 11 at 19:47

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  "I know that I would apply the sum formula" What is "the sum formula". "should I also be using this formula?" Isn't that formula the sum formula?
  $endgroup$
  – fleablood
  Mar 11 at 19:47
  

  
  
  
  

add a comment | 

$begingroup$

This question already has an answer here:

• 
  Sum of First $n$ Squares Equals $frac{n(n+1)(2n+1)}{6}$
  
  31 answers
  
  

How can I show that:$$sum_{k=1}^{n}({k^2})$$
Is equal to: $$frac{n(n+1)(2n+1)}{6}$$
I know that I would apply the sum formula, should I also be using this formula? $$sum_{k=1}^{n}k=frac{n(n+1)}2$$

discrete-mathematics summation

share|cite|improve this question

edited Mar 11 at 19:45

gt6989b

34.9k22557

asked Mar 11 at 19:32

Usama GhawjiUsama Ghawji

666

$endgroup$

share|cite|improve this question

edited Mar 11 at 19:45

gt6989b

34.9k22557

asked Mar 11 at 19:32

Usama GhawjiUsama Ghawji

666

share|cite|improve this question

edited Mar 11 at 19:45

gt6989b

34.9k22557

asked Mar 11 at 19:32

Usama GhawjiUsama Ghawji

666

edited Mar 11 at 19:45

gt6989b

34.9k22557

edited Mar 11 at 19:45

gt6989b

34.9k22557

asked Mar 11 at 19:32

Usama GhawjiUsama Ghawji

666

asked Mar 11 at 19:32

Usama GhawjiUsama Ghawji

666

1 Answer
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0

$begingroup$

Use proof by induction

1 - Demonstrate that it is true for $n=1$

2 - Demonstrate that if it is true for $n=k$ it must also be true for $n=k+1$

This comes down to demonstrating ...
$$ (k+1)^2+ frac{k(k+1)(2k+1)}{6} \= frac{(k+1)(k+2)(2(k+1)+1)}{6}$$

share|cite|improve this answer

answered Mar 11 at 19:50

WW1WW1

7,3401712

$endgroup$

add a comment | 

0

$begingroup$

Use proof by induction

1 - Demonstrate that it is true for $n=1$

2 - Demonstrate that if it is true for $n=k$ it must also be true for $n=k+1$

This comes down to demonstrating ...
$$ (k+1)^2+ frac{k(k+1)(2k+1)}{6} \= frac{(k+1)(k+2)(2(k+1)+1)}{6}$$

share|cite|improve this answer

answered Mar 11 at 19:50

WW1WW1

7,3401712

$endgroup$

add a comment | 

0

$begingroup$

Use proof by induction

1 - Demonstrate that it is true for $n=1$

2 - Demonstrate that if it is true for $n=k$ it must also be true for $n=k+1$

This comes down to demonstrating ...
$$ (k+1)^2+ frac{k(k+1)(2k+1)}{6} \= frac{(k+1)(k+2)(2(k+1)+1)}{6}$$

share|cite|improve this answer

answered Mar 11 at 19:50

WW1WW1

7,3401712

$endgroup$

add a comment | 

$begingroup$

Use proof by induction

1 - Demonstrate that it is true for $n=1$

2 - Demonstrate that if it is true for $n=k$ it must also be true for $n=k+1$

This comes down to demonstrating ...
$$ (k+1)^2+ frac{k(k+1)(2k+1)}{6} \= frac{(k+1)(k+2)(2(k+1)+1)}{6}$$

share|cite|improve this answer

answered Mar 11 at 19:50

WW1WW1

7,3401712

$endgroup$

share|cite|improve this answer

answered Mar 11 at 19:50

WW1WW1

7,3401712

answered Mar 11 at 19:50

WW1WW1

7,3401712

answered Mar 11 at 19:50

WW1WW1

7,3401712