By using induction, verify that each equation below is true for very positive integer n.

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

hello....somebody know how to solve this question? show me the clear step ,i am preparing for my test
thks a lot

Hello,

1² + 2² + 3² + ... + n² = n(n + 1)(2n + 1)/6

= = = = = = = = = = = = = = = = = = = = = = = = = = = = =
First step is to check this property is true for the first n
= = = = = = = = = = = = = = = = = = = = = = = = = = = = =
If n=1,
   1² = 1
   1×2×3/6 = 1
   Correct

If n=2,
   1² + 2² = 5
   2×3×5/6 = 5
   Correct

= = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Second step is to assume the property is true for a given n
= = = = = = = = = = = = = = = = = = = = = = = = = = = = =
1² + 2² + 3² + ... + n² = n(n + 1)(2n + 1)/6

= = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Third step is to prove it true for n+1
= = = = = = = = = = = = = = = = = = = = = = = = = = = = =
1² + 2² + 3² + ... + n² + (n + 1)²
   = n(n + 1)(2n + 1)/6 + (n + 1)² →→→ Since property is true for n
   = (n + 1)[n(2n + 1)/6 + (n + 1)] →→→ Factor out n+1
   = (n + 1)/6 × (2n² + n + 6n + 6) →→→ Factor out ⅙ and expand the rest
   = (n + 1)/6 × (2n² + 4n + 3n + 6) →→→ By grouping
   = (n + 1)/6 × [(2n(n + 2) + 3(n + 2)] →→→ Factor out n+2
   = (n + 1)/6 × (n + 2)(2n + 3)
   = (n + 1)[(n + 1) + 1][2(n + 1) + 1] / 6
Thus the property is proven for n+1

= = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Last step is to conclude
= = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Since:
   The property is true for the first ranks;
   Assuming it to be true for rank n will induce it to be true for rank n+1;
Then the property is true for all ranks.

Thus by mathematical induction have we proven that:
For any integer n:
1² + 2² + ... + n² = n(n + 1)(2n + 1)/6

Regards,
Dragon.Jade :-)

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