Prove ∀n≥1?

Question

1) Prove ∀n≥1
1^(2)+2^(2)+3^(2)+...+n^(2)= (n(n+1)(2 n+1))/(6);

Answer 1

Induction on n:

1) n=1 it's true
2) let it be true for n=k, let's prove this fact for n=k+1
Since we know 1^(2)+2^(2)+3^(2)+...+k^(2)= (k(k+1)(2 k+1))/(6), we obtain:
1^(2)+2^(2)+3^(2)+...+k^(2)+ (k+1)^2= (k(k+1)(2 k+1))/(6) + (k+1)^2 = [(k+1)(k(2k+1) + 6(k+1))]/6 = (k+1)[(2k^2+k + 6k+6)]/6 = (k+1)[(2k^2+7k+6)]/6 = (k+1)[(2k+3)*(k+2)]/6 =
[(k+1)*((k+1)+1)*(2(k+1)+1)]/6. Q E D

Answer 2

This is a classical example of the technique called "mathematical induction". It goes in two steps:
1. Check the relation for the start value (in your case, 1);
2. Prove that if the relation holds for any number, say n, it also holds for the following one, that is, n+1.
Then step 1 with step 2 will give you the proof for n=2; apply again and obtain for n=3, and so on. You don't have to do this explicitly, since step 2 applies to any n. The process can be continued indefinitely, so the relation is valid for all n values.

Example (simpler than your case):
1+2+3+...+n=n(n+1)/2 for any n>=1

Step 1: For n=1, the relation becomes simply 1=1*(1+1)/2, which is obviously true.
Step 2. We need to prove the relation for n+1, that is calculate
1+2+3+...+n+(n+1)
In this case, we know 1+2+...+n=n(n+1)/2, therefore we have
1+2+...+n+(n+1)=n(n+1)/2 + (n+1)
Factor out (n+1)/2, and you'll get
1+2+...+n+(n+1)=((n+1)/2)(n+2)=((n+1)/2)((n+1)+1)
where I also wrote n+2 as (n+1)+1, to pun n+1 in evidence. This is exactly your relation,with n replaced by n+1. To make it more obvious, replace n+1 with m:
1+2+...+(m-1)+m=(m/2)(m+1)=m(m+1)/2

For your relation,
step 1. Verify the relation for n=1; the calculations are elementary, you can do it.
step 2. Calculate the sum
1^(2)+2^(2)+...+n^(2)+(n+1)^(2)
by replacing the first terms of the sum with your relation; you'll get
1^(2)+2^(2)+...+n^(2)+(n+1)^(2)=(n(n+1)(2n+1))/6+(n+1)^(2)

The calculations from this point are simple, just factor the right hand side of the relation above. It may be easier to make the substitution above, n+1=m, or equivalent, n=m-1.

You;ll have the same relation for both n and m, except that m=n+1. Now you can finish your homework on your own.

Answer 3

If n=1
1^2 = 1*(1 + 1)*(1 + 2)/6 = 1*2*3/6 = 6/6 = 1

Let's assume it is true for n and prove for n+1:

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

= [n(n + 1)(2n + 1) + 6(n + 1)^2]/6 =
= (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)*2(n^2 + 3.5n + 3)/ 6 = (n + 1)*2(n + 1.5)(n + 2)/6 =
= (n + 1)(2n + 3)(n + 2)/6 = (n + 1)[2(n + 1) + 1][(n + 1) + 1]/6 =
= (n + 1)[(n + 1) + 1][2(n + 1) + 1]/6


QED

Answer 4

Make a recurence. It is true for n=1. Suppose that it is true for the rank n and prove the rank n+1; You just have to calculate : (n(n+1)(2 n+1))/(6) + (n+1)^2 and you must find. (n+1)(n+2)(2 (n+1)+1))/(6).

Answer 5

This isn't true for n = 1. Since 1! + 1 = 2 which is even. Let n = 2. Then 2! + 1 = 2 + 1 = 3 which is odd. Now let k! + 1 be odd. Need to show that (k+1)! + 1 is odd. Since k! + 1 is odd, we see that k! + 1 = 2m + 1 for some integer m. Then k! = 2m and so k! is even. Then (k+1)! = (k+1)(k!) = (k+1)(2m) = 2(k+1)m and so (k+1)! is even. Then (k+1)! + 1 = 2(k+1)m + 1 is odd. This completes the proof by induction.

Answer 6

So get yourself a large pile of small wooden cubes and a glue gun.
Glue together lots of different sized squares (1xnxn). Now make what I call "staggererd square pyramids" by stacking up squares of size 1,4,9,16, and 25 say. Do it in such a way so that corners line up, so you'll get something lopsided looking. Now make 6 of these staggered square pyramids.

It is a pleasant puzzle to fit them together into a solid block of size
5x6x11, which proves the formula in this case. Then just a little imagination shows you that you could do this same puzzle with 6 n-high pyramids, and get a solid block of size n(n+1)(2n+1), as the formula predicts.

Answer 7

suppose its true for n

sum [n+1] = n(n+1)(2n+1)/6 + (n+1)^2
= (n+1) [ 2n^2 + n + 6n+6 ] /6
= (n+1) [ 2n^2 + 7n + 6 ]/6
= (n+1) [(n+2) (2n+3) ]/6
so if true for n its true for any integer > n

sum[1] = 1*2*3/6 = 1 is true
therefore true for all positive integers