It is proved by mathematical induction that the nth power of 2 > 2n + 1 (n ∈ n *, n ≥ 3)

It is proved by mathematical induction that the nth power of 2 > 2n + 1 (n ∈ n *, n ≥ 3)

When n = 3, it is obviously true
If n = m, then 2 ^ m > 2m + 1
Then 2 ^ m * 2 ^ m > (2m + 1) * (2m + 1)
That is, 2 ^ (M + 1) > 4m ^ 2 + 4m + 1
However, 4m ^ 2 + 4m + 1 - (2 (M + 1) + 1) = 4m ^ 2 + 2m-2 > 0
That is, 2 ^ (M + 1) > 2 (M + 1) + 1
The proof is complete

It is proved by mathematical induction that the algebraic formula of multiplication on the left side of "from K to K + 1" when "(n + 1) (n + 2). (n + n) = 1 * 3 *... * (2n-1) * 2 ^ n" is Let n = k hold: (K + 1) (K + 2). (K + k) = 1 * 3 *... * (2k-1) * 2 ^ K Look at n = K + 1: left = [(K + 1) + 1] [(K + 1) + 2] [（k+1）+（k+1）] ＝[（k+1）（k+2）…… （k+k）]（k+1+k）（k+1+k+1）／（k+1） Shouldn't the two (n + n) places have the same N?... why can one be (K + 1) and the other be K (that is why there is (2k +!))

When n = k, it should be from [(K + 1) + 1] × [(K + 1) + 2] × [(K + 1) + 3] × [(K + 1) + 4],... " , until [(K + 1) + (K + 1)], then the last one in front of

Using mathematical induction to prove (n + 1) (n + 2) (n+n)=2^n*1*3*… *(2n-1) (n ∈ n +) left to add the algebraic expression is not proof!

It is proved that let a (n) = (n + 1) (n + 2) (n+n),B(n)=2^n*1*3*… *(2n-1).
When n = 1, a (1) = 1 + 1 = 2 = 2 ^ 1 * 1 = B (1),
When n = 2, a (2) = (1 + 2) (2 + 2) = 12 = 2 ^ 2 * 1 * 3 = B (2)
Suppose that when n = K (k > 2), a (k) = B (k) holds, that is, (K + 1) (K + 2) (k+k)=2^k*1*3*… *(2k-1) established,
Then when n = K + 1, a (K + 1) = (K + 1) (K + 2) = (K + 1) = (K + 2) = (K + 1) = (K + 1) = (K + 2) = (K + 1) = (k + 1) = (K + 1) = (K + (k+k)[k+(k+1)]=A(k)*[k+(k+1)]=A(k)*(2k+1)
B(k+1)=2^k*1*3*… *(2k-1)*[2(k+1)-1]=B(k)*[2(k+1)-1]=B(k)*(2k+1)
Obviously, a (K + 1) = B (K + 1) holds
So a (n) = B (n) holds for all n ∈ n +,
That is: (n + 1) (n + 2) (n+n)=2^n*1*3*… *(2n-1) established
The proof is over

Prove 1 + 2 + 3 + by mathematical induction +(n + 3) = (n + 3) (n + 4) / 2 (n ∈ R)_______

Note that (n + 3), when n equals 1, the maximum term is n + 3 = 4
Left 1 + 2 + 3 + 4 = 10
Right (1 + 3) (1 + 4) / 2 = 10
The left is equal to the right
So it was established

Prove equation 1 + 2 + 3 + by mathematical induction +(n+3)＝(n+3)(n+4) 2 (n ∈ n +), when n = 1 is verified in the first step, the item to be taken on the left is______

In equation 1 + 2 + 3 + +(n+3)＝(n+3)(n+4)
2 (n ∈ n +),
When n = 1, N + 3 = 4,
And the sum of continuous positive integers starting with 1 on the left side of the equation,
So when n = 1, the term on the left of the equation is: 1 + 2 + 3 + 4
So the answer is: 1 + 2 + 3 + 4

Using mathematical induction to prove "(n + 1) (n + 2) （n+n）=2n×1×3×… When × (2n-1), n ∈ n * ", and changing from" n = k "to" n = K + 1 ", the factor that should be multiplied on the left side is______ .

When n = K (K ∈ n *), the left formula is (K + 1) (K + 2) (K + k);
When n = K + 1, the left formula is (K + 1 + 1) · (K + 1 + 2) · · (K + 1 + k-1) · (K + 1 + k) · (K + 1 + K + 1),
Then the formula of multiplication on the left is (2k + 1) (2k + 2)
k+1=2（2k+1）．
So the answer is: 2 (2k + 1)

Using mathematical induction to prove (n + 1) (n + 2) （n+n）=2n•1•3•… In the case of (2n-1) (n ∈ n), for the proof from "K" to "K + 1", the algebraic formula to be added on the left is______ .

When n = k, the left is equal to (K + 1) (K + 2) （k+k）=（k+1）（k+2）… (2k), when n = K + 1, the left is equal to (K + 2) (K + 3) (K + k) (2k + 1) (2k + 2), so from "K" to "K + 1", the algebraic formula to be added on the left is (2k + 1) (2k + 2) (K + 1) = 2 (2k + 1)

Using mathematical induction to prove "(n + 1) (n + 2) （n+n）=2n×1×3×… When × (2n-1), n ∈ n * ", and changing from" n = k "to" n = K + 1 ", the factor that should be multiplied on the left side is______ .

When n = K (K ∈ n *), the left formula is (K + 1) (K + 2) (K + k);
When n = K + 1, the left formula is (K + 1 + 1) · (K + 1 + 2) · · (K + 1 + k-1) · (K + 1 + k) · (K + 1 + K + 1),
Then the formula of multiplication on the left is (2k + 1) (2k + 2)
k+1=2（2k+1）．
So the answer is: 2 (2k + 1)

1. Prove the inequality 1 + 1 / 2 + 1 / 3 +... 1 / (2 ^ n-1) > n / 2, deduce K + 1 from K The increase on the left is 2. In the sequence {an}, Prime Minister A1 > 0, common ratio Q > 0, and the sum of the first n terms is SN

1. The increasing formula on the left is 1 / 2 ^ k + 1 / (2 ^ k + 1) + 1 / (2 ^ k + 2) +. + 1 / (2 ^ k + 2 ^ K-2) + 1 / (2 ^ k + 2 ^ k-1),
That is, 1 / 2 ^ k + 1 / (2 ^ k + 1) + 1 / (2 ^ k + 2) +. + 1 / [2 ^ (K + 1) - 1]
2. Because each term is positive, the inequality to be proved is transformed into Sn * s (n + 2)

Using mathematical induction to prove inequality 1 / (n + 1) + 1 / (n + 2) + +1/(n+n)>13/24 From "proposition holds for n = k to proposition holds for n = K + 1" increases on the left side of inequality?

Added: 1 / (2k + 1) + 1 / (2k + 2) - 1 / (K + 1)
After the separation, the above formula is zero rain, so it holds