# Use induction to prove that for each natural number n, 1^2+2^2+3^2+...n^2=(n(n+1)(2n+1))/6?

Using induction prove for the natural numbers n

1^2+2^2+3^2+...n^2=(n(n+1)(2n+1))/6

### 8 Answers

- Scarlet ManukaLv 71 decade agoFavourite answer
Let P(n) be the statement "1^2 + ... + n^2 = n (n+1) (2n+1) / 6."

P(1) is the statement

1^2 = 1(2)(3) / 6

which is clearly true.

Now suppose P(n) is true for an integer n >= 1. Then P(n+1) is the statement

1^2 + ... + n^2 + (n+1)^2 = (n+1) (n+2) (2n+3) / 6

Now by P(n) we know

1^2 + ... + n^2 + (n+1)^2

= n (n+1) (2n+1) / 6 + (n+1) ^2

= (n+1) [n (2n+1) / 6 + (n+1)]

= (n+1) [n (2n+1) + 6(n+1)] / 6

= (n+1) (2n^2 + n + 6n + 6) / 6

= (n+1) (2n^2 + 7n + 6) / 6

= (n+1) (n+2) (2n+3) / 6

so P(n+1) is true.

Hence by induction P(n) is true for every positive integer n.

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- 4 years ago
Induction is a simple process. Don't get your brain in a knot staring at these equations you need to prove. Just jot down these three steps: 1. Plug in 1 for n into the equation, and show that both sides are equal 2. Next, take the equation as it was before (how you have it above) and write on the next term what it would be (in terms of n+1) 3. Replace the first n terms on the left with the right hand side formula. You will then have three terms in your equation, 2 on the left, one on the right. The rest is just algebra to simplify each side until you can show they are equal. If this seems confusing I'll do the first one for you, which might seem like really long but you'll learn it with practice. Remember that each of the terms on the left are actually just the last term plugging in the value for the value of n you are at. So in the first example, 4n is equal to 8 for n=2 . The first step means we are only showing the first term on the left hand side is equal to the formula on the right hand side for n=1, which is 4 on the left hand side. You want to be sure the first term is equal to the right hand side, 2n (n+1) so plug in one into the right hand side, which is 2*1(1+1) = 2*2 = 4. Easy. 4=4 . Next step. Now you want to show it is true no matter how big your n gets, so first assume the formula is true , that 4+8+12+...+4n = 2n(n+1) . We say you are assuming the formula is true up to the value n. But what about the next one, perhaps the formula is wrong? We need to check it will still be true for n+1. How do we do this? The next term in the left hand side of the formula would be 4(n+1). But then we have to change the right hand side, and replace n for n+1. Write this down to see what it looks like: 4+8+12+...+4n = 2n(n+1) (this formula is up to n) 4+8+12+...+4n + 4(n+1) = 2(n+1)(n+1 +1) (this formula we have went to the next term, n+1. simplifying the right hand side we get 4+8+12+...+4n + 4(n+1) = 2(n+1)(n+2) We must show the two sides are equal, so the trick is to now replace the original part of the left hand side with the formula we said to be true for n. 4+8+12+...+4n + 4(n+1) = 2(n+1)(n+2) turns into 2n(n+1) + 4(n+1) = 2(n+1)(n+2) See what I did there? Replace the first part of the formula with the original formula you are given, so this reduces into a simple algebra problem. Now you just need to multiply and add this out: 2n(n+1) + 4(n+1) = 2(n+1)(n+2) First Left hand side: 2n(n+1) + 4(n+1) = 2n^2 + 2n + 4n + 4 = 2n^2 + 6n + 4 Right hand side 2(n+1)(n+2) = 2 ( n^2 + n + 2n + 2) = 2 (n^2 + 3n + 2) = 2n^2 + 6n + 4 Both sides are equal and you are done. This might be sort of a mystery why you have proven this formula, at first. If you are still wondering about how it works, it helps to practice doing it and think about what you are doing. Show it is true for the first one. Then the second step shows you that no matter how big an n you use, the next term is always going to work in the formula. Thank you.

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- 1 decade ago
Using induction prove for the natural numbers n

1^2+2^2+3^2+...n^2=(n(n+1)(2n+1))/6

pf

if n=1 then clearly

1^2 = (1)(2)(3)/6 =1 so the statement is true when n=1.

Assume statement is true for n=k.

That is we assume

1^2+2^2+3^2+...k^2=(k(k+1)(2k+1))/6

Now, we show the statement is true for n=k+1.

1^2+2^2+3^2+...+k^2+(k+1)^2

by induction hypothesis we know that

1^2+2^2+3^2+...+k^2 = (k(k+1)(2k+1))/6

if we add (k+1)^2 to both sides we get:

1^2+2^2+3^2+...+k^2 + (k+1)^2 = (k(k+1)(2k+1))/6 + (k+1)^2

Now we simplify the right hand side.

(k(k+1)(2k+1))/6 + (k+1)^2 = (

k(k+1)(2k+1))/6 + 6(k+1)^2/6=

( k(k+1)(2k+1) + 6(k+1)^2 ) /6=

[ (k+1)(k(2k+1) + 6(k+1) ) ]/6=

[(k+1)(2k^2 +7k +6) ]/6=

[(k+1)(2k + 3)(k + 2) ]/6=

[(k+1)(2(k+1)+1)((k+1)+1)]/6

this shows the statement is true for n =k+1

so the statement is true for all n by induction.

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- UnknownDLv 61 decade ago
Let 1^1 + 2^2 + 3^2 + ... + n^2 = S

Show it's true for n = 1

1^2 = (1 * 2 * 3) / 6

1 = 6 / 6 = 1

Now prove it is true for n + 1.

S + (n + 1)^2 = (n(n + 1)(2n + 1)) / 6 + (n + 1)^2

What we're expecting is ((n + 1)(n + 2)(2n + 3)) / 6.

S + (n + 1)^2 = (n(n + 1)(2n + 1)) / 6 + 6(n + 1)^2 / 6

S + (n + 1)^2 = (n + 1) [ (n)(2n + 1) + 6(n + 1) ] / 6

S + (n + 1)^2 = (n + 1) [ 2n^2 + n + 6n + 6 ] / 6

S + (n + 1)^2 = (n + 1) [ 2n^2 + 7n + 6 ] / 6

S + (n + 1)^2 = [ 2n^3 + 7n^2 + 6n + 2n^2 + 7n + 6 ] / 6

S + (n + 1)^2 = [ 2n^3 + 9n^2 + 13n + 6 ] / 6

S + (n + 1)^2 = [ (n + 1)(n + 2)(2n + 3) ] / 6

Therefore the formula is true.

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- whitesox09Lv 71 decade ago
Base case n=1

1² = (1(1+1))(2(1)+1)) / 6

1 = 6/6

1 = 1

Assume true for n=k such that

1² + 2² + 3² + ... + k² = (k(k+1)(2k+1))/6

Show the statement holds for n=k+1

1² + 2² + 3² + .... + k² + (k+1)² = [(k+1)(k+2)(2(k+1)+1)]/6

Need to show that:

[(k+1)(k+2)(2(k+1)+1)]/6 - (k+1)^2 = (k(k+1)(2k+1))/6

[(k+1)(k+2)(2k+3)]/6 - (k+1)^2 = (k(k+1)(2k+1))/6

Divide both sides by (k+1):

[(k+2)(2k+3)]/6 - (k+1) = (k(2k+1))/6

(2k² + 7k + 6)/6 - (k+1) = (2k² + k)/6

Multiply both sides by 6:

(2k² + 7k + 6) - 6(k+1) = (2k²+ k)

2k² + 7k + 6 - 6k - 6 = 2k² + k

2k² + k = 2k² + k

Proven by mathematical induction.

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- PhiloLv 71 decade ago
well, for a starting point, if n = 2,

1² + 2² = 5 = 2(3)(5)/6

for continuation, suppose 1² + 2² + ... + k² = k(k+1)(2k+1)/6

we have to prove that

1² + 2² + ... + k² + (k+1)² = [k+1][(k+1) + 1][2(k+1) + 1]/6

so we take our supposition and substitute it into what we're proving to get

[k(k+1)(2k+1)/6 ] + (k+1)² = [k+1][(k+1) + 1][2(k+1) + 1]/6

and simplify both sides, separately,

[(k²+k)(2k+1)/6] + k² + 2k + 1 =? [k² + 2k + 1 + k + 1][2k + 2 + 1]/6

(2k³ + 3k² + k)/6 + k² + 2k + 1 =? (k² + 3k + 2)(2k + 3)/6

(2k³ + 3k² + k + 6k² + 12k + 6)/6 =? (2k³ + 9k² + 13k + 6)/6

(2k³ + 9k² + 13k + 6)/6 = (2k³ + 9k² + 13k + 6)/6

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- 1 decade ago
When you are proving any mathematical statement, you must manipulate one side (and only one side) to get the other side. Scarlet Manuka did that. That is what you MUST do with induction.

On scrap paper you can manipulate both sides to get an idea of what you must do, but not in a formal proof.

Source: Dr. Richard Libera, Professor of Mathematics, University of Delaware

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- 1 decade ago
Let n=1

1^1=1

1=1 good

Assume n=k

P(k)=1^2+2^2+...k^2=(k(k+1)(2k+1...

suppose n=k+1

P(k+1)=1^2+2^2+...k^2+ (k+1)^2=(k+1)((k+1)+1)(2(k+1)+1...

now subsitute...that should get you started

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