Determinant calculation: 123... N234... 1345... 2. N12... N-1 rt The first line is 1 to N, the second line is 2 to N, and then add a 1. The third line is n items starting from 3, and so on. The first item in the last line is n, and then 1 to n-1. Each line is composed of 1 to n

Determinant calculation: 123... N234... 1345... 2. N12... N-1 rt The first line is 1 to N, the second line is 2 to N, and then add a 1. The third line is n items starting from 3, and so on. The first item in the last line is n, and then 1 to n-1. Each line is composed of 1 to n

Except for the first row, subtract the I-1 row from the I row to get a new determinant, and then perform the previous operation on the new determinant, and finally get a better determinant (two sub diagonal elements are non-zero, and the rest are all zero except the first row)

1 2 2… 2 2 2 2… 2 2 2 3… 2 ………… 2 2 2… What is the determinant n?

Multiply the second row by - 1 and add it to the rest rows - 100... 0222... 2001... 0... 0000... N-2, multiply the second column by - 1 and add it to the rest columns - 100... 0020... 0001... 0... 0000... N-2 determinant = (- 1) 2 (n-1)! = - 2 (n-2)!

Calculating determinant

2 1 4 14 1 2 33 4 1 22 3 4 1 column 3 minus column 1 multiplied by 2, column 1 minus column 4 multiplied by 2, column 2 minus column 4 = 0 001-2 - 2 - 6 3-1 2 - 5 20 20 20 20 1 expand by line 1 = - 2 - 2 - 6-1 2 - 50 20 * (- 1) ^ (1 + 4) expand by line 3 = - 2 - 6-1 - 5 * 2 * (- 1) ^ (3 + 2) ^

Calculate the determinant 1 1 2 3 1 2 3 - 1 3 - 1 - 1 - 2 2 3 - 1 - 1
1 1 2 3
1 2 3 -1
3 -1 -1 -2
2 3 -1 -1
There is another question 2-1-1-1
1 3 0 0
1 0 4 0
1 0 0 5

1.
r2-r1,r3-3r1,r4-2r1
1 1 2 3
0 1 1 -4
0 -4 -7 -11
0 1 -5 -7
r3+4r2,r4-r2
1 1 2 3
0 1 1 -4
0 0 -3 -27
0 0 -6 -3
r4-2r3
1 1 2 3
0 1 1 -4
0 0 -3 -27
0 0 0 51
= -3*51 = -153.
2. This is a special "arrow" determinant, which has its specific solution
If you master this method, you will master the solution of this kind of determinant, especially for those with letter parameters
That is: use the elements on the main diagonal to change the elements corresponding to the first column to 0
So the determinant is transformed into upper triangular determinant
c1-(1/3)c2-(1/4)c3-(1/5)c4
k -1 -1 -1
0 3 0 0
0 0 4 0
0 0 0 5
k = 2+1/3+1/4+1/5
So determinant d = (2 + 1 / 3 + 1 / 4 + 1 / 5) * 3 * 4 * 5 = 167

Lim2 ^ (n + 1 + 1 / 2 + 1 / 3 +... + 1 / N) =?
lim2^(n+1+1/2+1/3+...+1/n)=?
Note: that is the (n + 1 + 1 / 2 + 1 / 3 +... + 1 / N) power of 2.

Are you wrong, 1 + 1 / 2 + 1 / 3 +... + 1 / N, which tends to infinity, so the limit is also infinity
The answer can't be equal to 2. If it is, there is a mistake in the book

1/1*2+1/2*3+...+1/n*(n+1)+...=( )
Seeking answers and solving ideas
Want a specific value
A、0 B、0.5 C、1 D、2
Thank you~

Idea: 1 / 1 * 2 = 1-1 / 21 / 2 * 3 = 1 / 2-1 / 31 / N * (n + 1) = 1 / n-1 / (n + 1) 1 / 1 * 2 + 1 / 2 * 3 +... + 1 / N * (n + 1) = 1-1 / 2 + 1 / 2-1 / 3 + 1 / n-1 / (n + 1) = 1 - 1 / (n + 1) = n / (n + 1) [Question supplement: want a specific value a, 0b, 0.5c, 1D, 2] when n

The sum of 1 / 1 + 1 / 2 + 1 / 3. + 1 / n

Use "Euler formula" (refer to related books): 1 + 1 / 2 + 1 / 3 + +1 / N = ln (n) + C, C is Euler constant, the value is 0.5772 .
Then 1 + 1 / 2 + 1 / 3 + 1 / 4 +... + 1 / 2007 + 1 / 2008 = ln (2008) + C = 8.1821 (about)

If M = {- 2,0,1}, n = {- 1,0,2}, then m ∩ n

｛0｝

Lim2 ^ X-1 / 4 ^ x + 1, X approaching positive infinity

Divide the numerator and denominator by 2 ^ x at the same time
obtain
Original limit
=LIM (x tends to positive infinity) (1-2 ^ - x) / (2 ^ x + 2 ^ - x)
Obviously, 2 ^ x tends to positive infinity and 2 ^ - x tends to zero
therefore
The numerator tends to constant 1 and the denominator to positive infinity
So the limit tends to zero

When n approaches infinity, lim2 ^ 1 / 2 + 1 / 4 + +How to find the exponent of 2 for 1 / 2 ^ n

When n approaches infinity, the sum of the first n terms s (n) = 1
So: lim2 ^ 1 / 2 + 1 / 4 + +1/2^n = lim2^1 = 2