Mathematical induction proves 1 2 3 4 。。。。。 n2 n4 n2 2, then n k 1

Is there a problem with the topic? 1＋2＋3＋4＋……n=n(n 1) 2 proves: When n 1, the left side of the equation = 1, and the right side of the equation = 1, assuming that n = k is true, i.e.,

1＋2＋3＋4＋……k=k(k 1) 2, then when n=k 1, the left side of the equation = 1 2 3 4 ......k (k 1) = k(k 1) 2 (k 1) = k(k 1) 2 2 (k 1) 2= (k 1) (k 2) 2, i.e. when n=k 1, the original is true.

The question is not difficult. When n=1, it is clearly true.

Assuming that n=k is true, there is: 1+2+......k 2=(k 4+k 2) 2 When n = k+1, left = 1 + 2 + ......k^2+(k^2+1)+(k^2+2)+(k^2+1)+…k^2+2k+1)

k 4+k 2) 2+(2k+1)k 2+(2k+1)(k+1) is arranged: left = (k 4+4k 3+7k 2+6k+2) 2 (next according to the perfect square formula, with the common factor, the content of the junior high school).

k^4+2k^3+k^2）+（2k^3+4k^2+2k）+（2k^2+4k+2）>/2

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

k+1)^4+(k+1)^2>/2

That is, when n=k+1, it is also true and proven.

n=4 2^4=16>=4^2=16

n=5 2^5=32>=5^2=25

n=6 2^6=64>=6^2=36

Hypothesis 2 n>=n2 holds for all n>=4.

That is, there are 2 k>=k 2 k>=4

2 (k+1)=2 k x 2>=k 2 x 2 because k 2 x2-(k+1) 2=k 2-2k-1 =(k-1) 2-2 >=3 2-2>=0

So 2 (k+1)>=(k+1) 2

So for all n>=4, there is 2 n>=n2

The first item 1 * 2 = 1 * 2 * 3 3 Establish early friends.

Suppose n=k 1*2+2*3+3*4+....+k(k+1)=1 3k(k+1)(k+2) holds.

Then when n=k+1, 1*2+2*3+3*4+....+k(k+1)+(k+1)(k+2)

1/3k(k+1)(k+2)+(k+1)(k+2)(k+1)(k+2)(1/3k+1)

1 3 (k+1) (k+2) (k+3) was established.

So 1*2+2*3+3*4+....+n(n+1)=1/3n(n+1)(n+2)

Give points to play mathematically, good sails, dates, hard work, and ......

Proof that when n 1, the left 1 and the right sail ridge (1 1) 1 2, there is obviously a left and a right, and the original inequality holds.

Suppose that when n k the original inequality holds the destruction of the sedan forest, i.e., 1 2 2 3 3 ......k k (k 1) k

Then when n k 1, left 1 2 2 3 3 ......k^k＋（k＋1）^（k＋1）

k＋1）^k＋（k＋1）^（k＋1）

k＋1）^k＋（k＋1）（k＋1）^k

1＋k＋1）（k＋1）^k

k＋2）（k＋1）^k

k＋2）（k＋2）^k

k＋2）^（k＋1）

Right (k 1 1) (k 1) (k 2) (k 1) is left and right, and the original inequality also holds.

To sum up, the original inequality holds.

Since n (n +1) = n (n +1) (n +2) - the first (n-1) n (n +1)] 3

So 2*1*2 3+. n (n +1).

1 * 2 * 3-0 2 * 3 * 4-1 * 2 * 3 + n（n +1）（n +2） -n-1），n（n +1）] 3

After the elimination, [Xiang Pai Dayun Item].

n（n +1）（n +2）] 3

So, 1 2 + 2 2 + 3 2 + n 2

n（n +1）（n +2）] 3 - n（n +1）] 2

n（n +1）[（n + 2）/ 3-1/2]

Or mathematics is modeled on the Channer method. Or.

3 - n-1）^ 3 = 2 * n ^ 2 +（n-1）^ 2-n

integer equations.

3-1 3 = 2 * (2 2 3 2 + n 2) + 1 2 +2 2 + n-1) 2] -2 3 4 + n).

3-1 = 2 *（1 ^ 2 2 ^ 2 ^ 3 2 + n ^ 2）-2 + 1 ^ 2 +2 ^ 2 + n-1）^ 2 + n ^ 2]-n ^ 2 - 2 +3 +4 + n）

3-1 = 3 *（1 ^ 2 +2 ^ 2 +3 ^ 2 + n ^ 2）-2-n ^ 2 - 1 +2 +3 + n）+1

3-1 = 3（1 ^ 2 +2 ^ 2 + n ^ 2）-1-n ^ 2-n（n +1）/ 2

3(1 2 +2 2 + n 2) = n 3 + n 2 + n(n +1) 2 = (n 2) (2n 2 +2 n + n +1) = n 2) dust beam (n +1) (2n +1).

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

When n=1, 1=1 2*1*(1+1) holds true When n=k-1 holds, i.e., 1+2+3+......k-1) with line = 1 2 * (k-1) * (pei key k-1 + 1) when n = k, 1 + 2 + 3 + ......k=1 2*(k-1)*(k-1+1)+k=1 2*(k-1)*k+k=1 2*(k+1)*k, so it is true to shout that no matter what the value of n is, 1+2+3+......n=1/2*n*（n...

Mathematical induction is when n=1 1*2=(1+1)(1+2) 3 holds.

When n=k 1*2+2*3+3*4+.k(k+1) = (k+1)(k+2) /3k

Then n=k+1 (k+1)(k+2) 3k+k(k+1) = (k+1+1)(k+1+2).

i.e. 1*2+2*3+3*4+...n(n+1)=1 3n(n+1)(n+2) holds.

ps: You didn't write the formula correctly (n+1)(n+2) should be placed on top of the score line, and the parentheses or something are added to understand that you are so easy to misunderstand.

Solution: 1-2 2+3 2-4 2+....+1)^(n-1)*n^2=(-1)^(n-1)*[n(n+1)]/2

1) When n=1, 1=(-1) 0 holds, i.e., when n=1, the above equation holds.

2) Suppose that when n=k (k is a positive natural number), the above equation is immediate.

1-2^2+3^2-4^2...1)^(k-1)*k^2=(-1)^(k-1)*k(k+1)/2

then when n=k+1.

1-2^2+3^2-4^2...1)^(k-1)*k^2+(-1)^k*（k+1）^2=(-1)^(k-1)*k(k+1)/2+(-1)^k*（k+1）^2=(-1)^k*(k+1)(k+2)/2

The above equation holds when n=k+1.

In summary, from (1)(2), we know 1-2 2+3 2-4 2....1)^(n-1)*n^2=(-1)^(n-1)*n(n+1)/2

It should be (-1) (n-1)*n*(n+1) 2.

Prove one and establish first.

Suppose for n terms and hold.

1-2^2+3^2-4^2+..1)^(n-1)*n^2=(-1)^(n-1)*n*(n+1)/2

N+1 term sum.

1-2^2+3^2-4^2+..1)^(n-1)*n^2+(-1)^(n)*(n+1)^2

-1)^(n-1)*n*(n+1)/2+(-1)^(n)*(n+1)^2

-1)^(n)*(n+1)^2-(-1)^(n)*n*(n+1)/2

-1)^(n)*[n+1)^2-n*(n+1)/2]

-1)^(n)*[n+1)(n+1-n/2)]

-1)^(n)*(n+1)(n+2)/2

This equation also holds for the sum of n+1 terms, which is proved by mathematical induction to obtain 1-2 2+3 2-4 2+...1) (n-1)*n 2=(-1) (n-1)*n*(n+1) 2 holds for all positive integers n.