Prove 2n 18 n 2 8n 8 by mathematical induction

There is no condition for n given in the problem, assuming that n is a positive integer.

2n-18=4), i.e., 2k-180 when n=k+1 (k>=4), (k+1) 2-8(k+1)+8)-(2*(k+1)-18).

k^2+2k+2-8k-8+8)-(2k+2-18)k^2+2k+2-8k-2k+16

k^2-8k+18

k^2-10k+2k+26-8

k^2-10k+26+2k-8

2k-8 (k^2-10k+26>0)

Because k>=4, 2k-8>=0

So ((k+1) 2-8(k+1)+8)-(2*(k+1)-18)>0

i.e. 2*(k+1)-18 < k+1) 2-8(k+1)+8 Therefore, when n=k+1 (k>=4), the inequality holds.

In summary, 2n-18

The inequality is true when n=1 is proved first, and then it is true when n=k is assumed, and the hypothesis is proved when n=k+1 is true.

When n=1 is 2*1-18=-16 and 1 -8*1+8=1, the inequality is obviously true.

n=2 at ......

n=3……n=4……

n=5……Suppose that when n=k(k>=5), 2k-8+1>=3 when 2k-18=5, there is 2<2k-8+1

2k-18 again, so 2k-18+2

Proof: 2(n-9)<(n-4) 2

When n=0, the -18<16 proposition holds.

When n=1, the -16<9 proposition holds.

When n=9, the proposition 0<25 holds.

Suppose that the proposition is true when n=k (k n0, k is a natural number) and also when n=k+1 2(k-8)<(k-3) 2 is also true as above.

Hence 2n-18

sn=n(n+1)(2n+1)/6。

The process is as follows:

an = n²

sn = 1² +2² +3² +n² =n(n+1)(2n+1)/6

Inductive proof:

n = 1, 1 (1+1) (2 1+1) 6 = 6 6 = 1, the sum formula is correct.

When n = k, sk = 1 +2 +3 +k = k(k+1)(2k+1) 6 holds.

s(k+1) =k(k+1)(2k+1)/6+(k+1)²

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

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

k+1)[2k²+7k+6]/6

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

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

Proven. Additional Information:

1）（a+b）³＝a³＋3a²b＋3ab²＋b³

2）a³+b³=a³+a²b-a²b+b³=a²（a+b）-b（a²-b²）=a²（a+b）-b（a+b）（a-b）

a+b)[a-b(a-b)]=a+b) hall withered (a-ab+b).

3) a -b = a -a b + a marketing b-b = a (a-b) + b (a -b) = a (a-b) + b (a + b) (a-b).

a-b）[a²+b（a+b）]=a-b）（a²+ab+b²）

4）（a-b）³＝a³－3a²b＋3ab²-b³

a-b)³=a-b)(a-b)²=a-b)(a²-2ab+b²)=a³-3a²b+3ab²-b³

The most simple and common mathematical induction.

is to prove that when n is equal to any natural number.

A certain proposition is true. The proof is in two steps:

1. Prove that the proposition is true when n = 1.

2. Assuming that the proposition is true when n=m, then it can be deduced that the proposition is also true when n=m+1. (m stands for any natural number).

The principle of this method is to first prove that the proposition is true at a certain starting value, and then to prove that the process from one value to the next is valid. When both of these points have been proven, then any value can be deduced by using this method repeatedly.

Think of this approach as a domino effect.

Maybe it's easier to understand.

When Zheng n=1, the left finch is dismantled = 1 + 2 * 1 = 2 and the right is = 1 * (1 * 2 + 1) = 3

The equation holds. When n=k is assumed, the equation holds.

i.e. 1+2+...2k=k(2k+1)

Then when n=k+1.

Left = 1+2+.2n+(2n+1)+(2n+2)(1+2n+2)*(2n+2)/2

n+1)*(2n+3)

The equation also holds.

n=1 Left Burning Empty Side = 1 6 Right = 1 6 Formation n=2 Left Side = 1 4 Right Ming Banquet Side = 2 9 Formation Setting Setting N=k (k>=2) True.

1 2-1 (k+2)=2 1 ((k+2)(k+3)),2,n=1 left =1 6 right =1 6 true.

n=2 Left=1 4=9 36 Right=2 9 =8 36 Left>> Right.

The inequality does not hold.

For example, 1 2-1 (n+2) (n+2) 18, then n (n+2) (n+2) 9

n-1)(n-4)≥0

n 4 or n 1,2,

When n=1, the left = 1 2 = 1 right = 1 * (1 + 1) * (2 + 1) The slag rent key is 6 = 1; Let n=k become immediate: Liang Ye 1 2+2 2+....+k 2=k(k+1)(2k+1) 6 then 1 2+2 2+....+k^2+(k+1)^2=k(k+1)(2k+1)/6+(k^2+2k+1)=(2k^3+3k^2+k+6k^2+12k+6)/6=(k+1)(k+2)(2k+3)/6=(k+1)[(k+1)+1...

n=2 ((n+1)/2)^n= [2+1)/2]^2= n!=2*1=2 So((n+1) 2) n> n!Establish.

n>2 Suppose n=k holds the original formula, i.e., ((k+1) 2) k> k!i.e. (k+1) k 2 k>k!(1)

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

Cause(k+2) (k+1)>2(k+1) (k+1).3)

3) Substituting (2) (k+1+1) 2) (k+1)=(k+2) (k+1) (2*2 k)>2(k+1) (k+1) (2*2 k)=(k+1) (k+1) 2 k=(k+1)*(k+1) k 2 k(4)

Substituting (1) into (4) gives ((k+1+1) 2) (k+1)>(k+1)*k!=(k+1)!i.e. n=k+1((n+1) 2) n > n!Establish.

2 n-1) stool slip judgment (2 n+1)>n (n ten1) (n 3, n jujube to n+), 1-2 (2 n+1)> 1-1 (n+1), 2 (2 n+1).

Simply write: n=1 2+2>1 is true.

This is true if n=k.

2^k+2>k^2

n=k+12^(k+1)+2-(k+1)^2=2*2^k+2-k^2+2k+1

2^k+(2^k+2-k^2)+2k+1 k>02^k+2>k^2

So 2 (k+1)+2-(k+1) 2>0 so for k+1 also holds. Proven.