Prove 1 / 2 + 2 / 2 ^ 2 + 3 / 3 ^ 2 + by mathematical induction +N/2^n=2- (n+2) /2^n when n=k+1, the left end is added to the left end of n+k

prove
N = 1, left = 1 / 2, right = 2-3 / 2 = 1 / 2
Left = right
Suppose n = K
1 / 2 + 2 / 2 ^ 2 +... + K / 2 ^ k = 2 - (K + 2) / 2 ^ k holds
that
When n = K + 1
1/2+2/2^2+...+k/2^k+(k+1)/2^(k+1)
=2-(k+2)/2^k+(k+1)/2^(k+1)
=2-2(k+2)/2^(k+1)+(k+1)/2^(k+1)
=2-[2(k+2)-(k+1)]/2^(k+1)
=2-(k+1+2)/2^(k+1)
Left = right
The equation holds
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Prove 1 + 2 + 3 + by mathematical induction +n2=n4+n2 2, then when n = K + 1, the left end should be added on the basis of n = K______ .

When n = k, the left end of the equation = 1 + 2 + +k2，
When n = K + 1, the left end of the equation = 1 + 2 + +k2+（k2+1）+（k2+2）+（k2+3）+… +(K + 1) 2, added 2K + 1 term, that is (K2 + 1) + (K2 + 2) + (K2 + 3) + +（k+1）2
So the answer is: (K2 + 1) + (K2 + 2) + (K2 + 3) + +（k+1）2

It is proved by mathematical induction that the inequality 1 / 2 ^ 2 + 1 / 3 ^ 3 +... + 1 / N ^ n < (n-1) / N holds for any positive integer n greater than 1

When n = 2, left is 1 / 2 ^ 2, right is 1 / 2, left is < right
Suppose n = k holds, that is, 1 / 2 ^ 2 + 1 / 3 ^ 3 +... + 1 / K ^ k < (k-1) / K
When n = K + 1,1 / 2 ^ 2 + 1 / 3 ^ 3 +... + 1 / K ^ k + 1 / (K + 1) ^ (K + 1) < (k-1) / K + 1 / (K + 1) ^ (K + 1) < (k-1) / K + 1 / (K + 1) ^ 2 < (k-1) / K + 1 / (K + 1) k = K / (K + 1), it holds for K + 1
According to induction, it holds for any n greater than 1

Using mathematical induction to prove inequality: 1 + 1 / √ 2 + 1 / √ 3 + + 1 / √ n ＞ √ (n + 1) (n ＞ = 3 and N ∈ n *)

And then it proved out~

Proving inequality 2 ^ n by mathematical induction

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I only write the main part
Suppose 1 / (2 * 2) + 1 / (3 * 3). + 1 / (n * n)

It is proved that the following inequality holds for all positive integers n greater than 2 +n)(1+ 1/2 + 1/3 +… + 1/n) ≥ n^2+n-1

prove:
Let f (n) = (1 + 2 + 3 +) +n)(1+ 1/2 + 1/3 +… + 1/n)-n^2-n+1
f(3)=(1+2+3)(1+ 1/2 + 1/3)-9-3+1=6*11/6-9-3+1=0
f(n+1)-f(n)=(1+2+3+… +n+n+1)[1+ 1/2 + 1/3 +… + 1/n+1/(n+1)]-(n+1)^2-n
-(1+2+3+… +n)(1+ 1/2 + 1/3 +… + 1/n)+n^2+n-1
=1+(n+1)(1+ 1/2 + 1/3 +… + 1/n)+(1+2+3+… +n)(n+1)-2n-2
>1+n+1+(n+1)^2-2n-2>0
F (n) increases monotonically
f(n)>f(3)≥0

Using mathematical induction to prove Equality: n ∈ n, n ≥ 1, 1 − 1 2+1 3−1 4+… +1 2n−1−1 2n＝1 n+1+1 n+2+… +1 2n．

It is proved that: (1) when n = 1, left = 1 − 12 = 12 = right, the equation holds. (2) if n = k, the equation holds, that is, 1 − 12 + 13 − 14 + +12k−1−12k＝1k+1+1k+2+… +At 12K, 1 − 12 + 13 − 14 + +12k−1−12k+(12k+1−12k+2)=1k+1+1k+2+… +12k+(12k+1−12k+2)=1k+...

It is known that the positive number ABC is the length of the three sides of a triangle, and the equation a ^ 2 + B ^ 2 + C ^ 2 = AB + AC = BC holds. Try to determine the shape of the triangle and explain the reason

Multiply the original formula by 2 to get 2A ^ 2 + 2B ^ 2 + 2C ^ 2-2ab-2ac-2bc = 0
(a-b) ^ 2 + (A-C) ^ 2 + (B-C) ^ 2 = 0
So a = b = c. a triangle is an equilateral triangle

It is known that the a power of 2 is 3, the B power of 2 is 6, and the C power of 2 is 12. Find the relationship between ABC

Because 2 ^ B = 6
So (2^b) ^2=6^2=36=3*12
2 ^ (2b) = 3 * 12
And 2 ^ a = 3,2 ^ C = 12
SO 2 ^ (2b) = (2 ^ a) * (2 ^ C)
2 ^ (2b) = 2 ^ (a + C)
So 2B = a + C

Prove 1 / 2 + 2 / 2 ^ 2 + 3 / 3 ^ 2 + by mathematical induction +N/2^n=2- (n+2) /2^n when n=k+1, the left end is added to the left end of n+k