Two arithmetic sequences {an}, {BN}, (a1 + A2 +...) +an）/（b1+b2+… BN) = (7n + 2) / (n + 3), then A5 / B5 = () What I want to ask is why this topic can't be done like this: Let Sn = a1 + A2 + an Tn=b1+b2+… bn Then the molecule = Sn = 7n + 2 Denominator = TN = n + 3 a5=S5-S4=37-30=7 b5=T5-T4=8-7=1 a5/b5=7/1=7 Why is it wrong to do so? I'll see it tomorrow

This is obviously wrong, like you say Sn / TN = (7n + 2) / (n + 3) S5 / T5 = 37 / 8s4 / T4 = 30 / 7, I can say S5 = 37, T5 = 8s4 = 30, T4 = 7, of course, it can also say S5 = 74, T5 = 16s4 = 30, T4 = 14, at this time, the result is obviously different. This question should be done as follows: S9 = 9 (a1 + A9) / 2 T9 = 9 (B1 + B9)

Given an = 2n + 2 ^ n, find the value of Sn = a1 + A2 + a3 +... + an

Because an = 2n + 2 ^ n
So: SN = a1 + A2 + a3 +. + an
=2(1+2+3+...+n)+2^1+2^2+2^3+...+2^n
=2x（1+n）/2+(2-2^n x2)/(1-2)
=2n+2+2^nx2-2
=2^(n+1)+2n
Analysis: this problem can be solved by group summation

Determinant calculation: - A1 A1 0 0 0 0 -a2 a2…… 0 0 …… …… 0 0 0…… -an an 1 1 1 1 1

-A1 A1 0... 00 - A2 A2... 00... - an an1 1... 11 add 2,3,..., n columns to the first column to get 0 A1 0... 00 - A2 A2... 00... - an ANN + 1 1 1... 11 expand the determinant according to the first column = (n + 1) * - 1) ^ n * A1 A2

The determinant A001 1； a1 1 0 1…… 1；a2 1 1 0…… 1；：：：an 1 1 1…… 0 is the wrong way to find the sum of the algebraic cofactors of A0 and the value of the determinant

a0 0 1 1…… one
a1 1 0 1…… one
a2 1 1 0…… one
；：：：
an 1 1 1…… 0
It's not a determinant! You can see where the main diagonal is
For an element, we can only find its algebraic cofactor, but not the sum of its algebraic cofactor

Calculate the determinant | 1 + A1 1 1 | | 1 1+a2 …… 1 | | …… …… …… | | 1 1 …… 1+an |

1+a1 1 1 …… 1 1
1 1+a2 1 …… 1 1
1 1 1+a3 …… 1 1
…… …… …… …… …… …… ……
1 1 1 …… 1+a 1
1 1 1 …… 1 1+an
Subtract line n-1 from line n, subtract line n-2 from line n-1 The first line is subtracted from the second line
1+a1 1 1 …… 1 1
-a1 a2 0 …… 0 0
0 -a2 a3 …… 0 0
…… …… …… …… …… …… ……
0 0 0 …… a 0
0 0 0 …… -a an
Click on the first line to expand
The original formula = (1 + A1) * A2 * A3 * A4 * *a*an
＋ (-1)*(-a1)*a3*a4*…… *a*an
＋ [(-1)^2]*(-a1)*(-a2)*a4*…… *a*an
＋………… ＋
＋ [(-1)^(n-2)]*(-a1)*(-a2)*(-a3)*…… *(-a)*an
＋ [(-1)^(n-1)]*(-a1)*(-a2)*(-a3)*…… *(-a)*(-a)
= a1*a2*a3*a4*…… *a*an
＋a2*a3*a4*…… *a*an
＋ a1*a3*a4*…… *a*an
＋ a1*a2*a4*…… *a*an
＋………… ＋
＋ a1*a2*a3*…… *a*an
＋ a1*a2*a3*…… *a*a
(1) If the sequence A1, A2, A3 If at least two numbers in an are equal to zero, then
There will be at least two identical rows of 1 in the determinant,
The original determinant = 0
(2) If the sequence A1, A2, A3 If a i = 0 (where I ∈ [1, n]), then
The original determinant = A1 * A2 * A3 * * ai-1 * ai+1*…… *An (the product of the remaining n-1 numbers in a sequence without AI)
(3) If the sequence A1, A2, A3 And an are not zero
The original determinant = A1 * A2 * A3 * A4 * *a*an * [ 1+1/a1+1/a2+1/a3+…… +1/an ]

5 x-4 under the root of LIM X-1

Original formula = LIM (x-1) [√ (5x-4) + √ x] (5x-4-x)
=4lim (x - > 1) [√ (5x-4) + √ x] 1 / 2
=1 / 4 × (1 + 1)
=2

The limit of xsinx / (x ^ 2 + 1) when x tends to infinity

The above answer is obviously wrong and should be divided into three parts
x/(x^2+1) * sinx
The limit of X / (x ^ 2 + 1) when x tends to infinity is 0
And SiNx is bounded, so the limit is 0

Solving problems of lobita's rule: LIM (x approaches 0) 1 / X multiplied by arcsinx
Another problem is LIM (x approaches 0) (x power of E - x power of E) / X

Finding LIM (x → 0) ln (arcsinx) / Cotx

Using the law of lobida
Original formula = LIM (x → 0) - Sin & # 178; X / arcsinx √ (1-x & # 178;)
Let sin t = X
The original formula = LIM (t → 0) - (Sin & # 178; (Sint)) / T
Once more the law of lobita
The original formula is LIM (t → 0) - 2Sin (Sint) * cos (Sint) * cost = 0

Finding x tends to 0, Lim xcos (1 / x)

When x tends to zero, X is an infinitesimal and COS (1 / x) is a bounded function
So: Lim xcos (1 / x) = 0