Determinant calculation | x + A1 A2 A3 an| |x x+a2 a3 …… an| …… |a1 a2 a3 …… x+an|

Determinant calculation | x + A1 A2 A3 an| |x x+a2 a3 …… an| …… |a1 a2 a3 …… x+an|


This irregularity:
x+a1 a2 a3 .an
x x+a2 a3 .an
.
a1 a2 a3 .x+an
Either (1,2) position is A1, or (n, 1) position is X
I can't contact you (anonymous), if you have any questions, please contact me or ask me!



Given (3x-1) 7 = a7x7 + a6x6 + a5x5 + a4x4 + a3x3 + a2x2 + a1x + A0, find the value of A7 + A6 + A5 + A4 + a3 + A2 + a1x + A0
The 7th power of (3x minus 1) is equal to the 7th power of a7x plus the 6th power of a6x plus the 5th power of a5x plus the 4th power of a4x plus the 3rd power of a3x plus the 2nd power of a2x plus a1x plus A0. Find the value of A7 + A6 + A5 + A4 + A3 + A2 + A1 + A0


Any power of 1 is equal to 1, so the value of A7 + A6 + A5 + A4 + a3 + A2 + a1x + A0 is required
(3x-1) ^ 7 = the X in A7 * x ^ 7 + A6 * x ^ 6 + A5 * x ^ 5 + A4 * x ^ 4 + a3 * x ^ 3 + A2 * x ^ 2 + A1 * x + A0 is equal to 1, that is:
(3*1-1)^7
=a7*1^7+a6*1^6+a5*1^5 +a4*1^4 +a3*1^3+a2*1^2+a1*1+a0
=a7+a6+a5 +a4 +a3+a2+a1x+a0
=2^7
=128



(X & sup2; - x + 1) 4 = A0 + a1x + a2x2 + a3x3 + a4x4 + a5x5 + a6x6 + a7x7 + X8, then A0 + A1 + A2 + a3 + A4 + A5 + A6 + A7 + A8 =?


Let x = 1
Then any power of X is 1
So the right side = A0 + A1 + A2 + a3 + A4 + A5 + A6 + A7 + A8
=(1²-1+1)4
=1



In the sequence {an}, the adjacent two terms an and a (n + 1) are two parts of the corresponding quadratic equation x ^ 2 + 3nx + BX = 0
Two adjacent terms in sequence {an} an.an +1 is two of the equation x ^ 2 + 3nx + BN = 0. If A1 = 2, try to find the value of B100


(an)+(an+1)=-3n (an+1)+(an+2)=-3n-3
We get (an + 2) - (an) = - 3
Because A1 = 2, A101 = 2 + 50 * (- 3) = - 148
Because A2 = - 5, A100 = - 5 + 49 * (- 3) = - 152
B100 = A100 * A101 = 22496
Merlin's? Sophomore's?