In the equal ratio sequence {an}, for any natural number n, there are a1 + A2 + +An = 2 ^ n-1, then (A1) ^ 2 + (A2) ^ 2 + +（an）^2=?

In the equal ratio sequence {an}, for any natural number n, there are a1 + A2 + +An = 2 ^ n-1, then (A1) ^ 2 + (A2) ^ 2 + +（an）^2=?

In the equal ratio sequence {an}, for any natural number n, there are a1 + A2 + +An = 2 ^ n-1 means Sn = 2 ^ n-1, so an = sn-sn-1 = (2 ^ n-1) - [2 ^ (n-1) - 1] = 2 ^ (n-1) so (an) ^ 2 = [2 ^ (n-1)] ^ 2 = 4 ^ (n-1) means {(an) ^ 2} is an equal ratio sequence with 1 as the first term and 4 as the common ratio, so (A1) ^ 2 + (A2)
Let A1, A2 It is proved that any n-dimensional vector can be expressed linearly by them
Let a be any n-dimensional vector, because A1, A2 , an, a are n + 1 n-dimensional vectors, so A1, A2 And because A1, A2 So r (A1, A2 ，an，a）=r（a1，a2，… , an) = n, so a can be determined by A1, A2 And the expression is unique
Triangle ABC, D is the point on BC, BD = 3DC, G is the midpoint of AD, the extension line of BG intersects AC and E, then BG: Ge
Come on!
Let BG = x, Ge = y,
Then be = x + y
Do DF parallel to be and AC parallel to F
So in the triangle ADF, Ge: DF = Ag: ad = 1:2, so DF = 2Y
In triangle CBE, DF: be = DC: BC = 1:4, that is, 2Y / (x + y) = 1 / 4
The solution is X: y = 7:1
That is BG: Ge = 7:1
F (x) = x ^ k sin1 / X (x ≠ 0), 0 (x = 0) when k satisfies what conditions, the function is differentiable when x = 0;
limx->0 x^ksin1/x
=limx->0 x^(k+1)(sin1/x)/(1/x)=x^(k+1)
If there is a limit at x = 0 and the left and right limits are equal, then k > = - 1
It is proved that the sum of the first number sequence {an-sn} belongs to the natural sequence {an-sn}
∵ Sn = n-an, ∵ a (n + 1) = s (n + 1) - S (n) = (n + 1) - A (n + 1) - N + a (n) = 1 + a (n) - A (n + 1); ∵ 2A (n + 1) = 1 + a (n); ∵ 2A (n + 1) - 2 = 1 + a (n) - 2, namely: 2 [a (n + 1) - 1] = a (n) - 1; ∵ a (n + 1) - 1] / [a (n) - 1] = 1 / 2; ∵ an-1} is an equal ratio sequence
Let A1, A2, --, as be linearly independent, then B1, B2, BS be linearly independent if and only if
A. Two vector groups are equivalent. B, matrix A = (A1, A2, an) is equivalent to matrix B = (B1, B2, BS). Why B
A no!
For example:
a1=(1,0,0),a2 =(0,1,0)
b1=(0,2,0),b2=(0,0,1)
The two vector groups are linearly independent, but they are not equivalent, and no one can represent who
B correct
Because a and B are equivalent, that is, a can be transformed into B by elementary transformation
Elementary transformation does not change the rank of matrix and the rank of column
So a and B are equivalent, which means that the rank of a and B is equal, that is, the rank of two vector groups is the same
So r (b) = R (a) = s, so B1, B2, BS are linearly independent
On the contrary, the two vector groups are linearly independent, and the number of vector groups is the same
So r (a) = R (b) = s
So a and B are equivalent
As shown in the figure, in △ ABC, D is the midpoint of AC side, AE ‖ BC, ed intersects AB at point G, and the extension line of intersecting BC is at point F, if BG: GA = 3:1, BC = 8
Let AE = CF = x, then BF = 8 + X. ∵ AE ∥ BC ∵ AEG ∥ BGF ∥ aebf = agbg = 13, that is, X8 + x = 13. The solution is: x = 4, that is, the length of AE is 4
Analysis of function discontinuity type and supplement the definition of removable discontinuity f (x) = (1 + 2x) ^ 1 / X
X = 0 is the break point
lim (1+2x)^(1/x)
=lim[(1+2x)^(1/(2x))]^2
=e^2
So x = 0 is a removable breakpoint
Supplementary definition f (0) = e ^ 2
1. Let the sum of the first n terms of the sequence {an} be Sn = 2n ^ 2, {BN} be an equal ratio sequence, and A1 = B1, B2 (A2 -- A1) = B1, let CN = an / BN to find the first n terms and TN of the sequence {cn}
The general terms of an and BN have been obtained, an = 4n-2, BN = 2 × () n-1, CN = (2n-1) × 4 (n-1)
How to ask for the back?
Tn=-11/9+(2/3×n-1/9)×4^nTn=（2-1）×4^（1-1)+（2×2-1）×4^(2-1)…… (2n-1) × 4 ^ (n-1) is the formula 4tn = (2-1) × 4 ^ (2-1) + (2 × 2-1) × 4 ^ (2-1) If (2n-1) × 4 ^ n is formula (2), then (1) - 2 obtains - 3tn = 1-2 × 4 ^ 1-2 × 4 ^ 2 - -2×4^...
Cn=(2n-1)*4^(n-1) =2n*4^(n-1)-4^(n-1)=n*2^(2n-1)-4^(n-1)
It is divided into two sequences: DN = n * 2 ^ (2n-1), FN = 4 ^ (n-1), and the sum of the first n terms is calculated respectively
The first n terms of DN and:
S＝1*2^1+2*2^3+3*2^5+…… +n*2^(2n-1)
4S＝1*2^3+2*2^5+3*2^7+…… +(n-1) * 2 ^ (2n-1) + n * 2 ^ (2... Expansion)
Cn=(2n-1)*4^(n-1) =2n*4^(n-1)-4^(n-1)=n*2^(2n-1)-4^(n-1)
It is divided into two sequences: DN = n * 2 ^ (2n-1), FN = 4 ^ (n-1), and the sum of the first n terms is calculated respectively
The first n terms of DN and:
S＝1*2^1+2*2^3+3*2^5+…… +n*2^(2n-1)
4S＝1*2^3+2*2^5+3*2^7+…… +(n-1)*2^(2n-1)+n*2^(2n+1)
By subtraction, 3S = - [2 ^ 1 + 2 ^ 3 + 2 ^ 5 + 2 ^ 7 + +2^(2n-1)]+n*2^(2n+1)
S＝－2*(4^n-1)/9 + [n*2^(2n+1)]/3
(1-4 ^ n) / 3
The sum of the first n terms of CN: - 2 * (4 ^ n-1) / 9 + [n * 2 ^ (2n + 1)] / 3 + (1-4 ^ n) / 3
Let A1, A2, A3... An be a group of n-dimensional vectors. It is proved that the N vectors are linearly independent if and only if any n
Let A1, A2, A3... An be a group of n-dimensional vectors. It is proved that these n-dimensional vectors are linearly independent if and only if any n-dimensional vector can be expressed linearly by them
Necessary condition: any (n + 1) n-dimensional vector must be linearly related, that is, any n-dimensional vector B can be expressed linearly by A1, A2, A3... An
Sufficient condition: obviously