If X 5 2 X 3 1 a0 a1 X 1 a2 X 1 2 a5 X 1 5 holds any real number X, then a3

If you first take the value of x as 1, then the left and right sides of the equation become: 1+2+1=a0+0+0+0+0+0, so a0=4, and you take the value of x to 0, then the equation becomes: 0+0+1=a0-a1+a2-a3+a4-a5, that is:

1=4-a1+a2-a3+a4-a5, then a1-a2+a3-a4+a5=3, listed as equation (1), you take the value of x as 2, then the equation becomes: 32+16+1=a0+a1+a2+a3+a4+a5, that is: 49=4+a1+a2+a3+a4+a5, then a1+a2+a3+a4+a5=45, as equation (2), you add equation (1) and equation (2) and divide by 2, you get:

a1+a3+a5=24, and the coefficient of x to the power of 5 is a5, which is compared with the corresponding coefficient on the left side of the equation: a5=1, then a1+a3=23Later, you can find the same reason and continue to calculate to get the value of A3!

x-1)^n=∑c(n,i)x^i*(-1)^(n-i)

So right. The term x 5 is a5*c(5,5)x 5*(-1) (5-5)=a5*x 5, and the coefficient is a5, a5=1;

The term for x 4 is a5*c(5,4) x 4*(-1) (5-4)+a4*c(4,4) x 4*(-1) (4-4).

a5*5*x^4*(-1)+a4*1*x^4*1

5a5+a4) x 4, the coefficient is -5a5+a4=0, a4=5a5=5;

The term for x 3 is a5*c(5,3) x 3*(-1) (5-3)+a4*c(4,3) x 3*(-1) (4-3)+a3*c(3,3) x 3*(-1) (3-3).

a5*c(5,3)x^3-a4*c(4,3)x^3+a3*x^3

10a5-4a4+a3]x^3

10+a3] x 3, the coefficient is -10+a3=2, a3=12.

Analysis: From the meaning of the question to x=t+1, then:

t+1)^5

2*(t+1)³

1=a0+a1*t+a2*t²

a3*t³a4*t^4

a5*t^5

Then it is easy to know that a3 is (t+1) 5

2*(t+1)³

1 with a coefficient containing t.

The term containing t in binomial equation (t+1) 5 is c(5,2)*t = 10t, and the term containing t in t+1) is t, that is, the term containing t in 2*(t+1) is 2t

So we know (t+1) 5

2*(t+1)³

The term containing t in 1 is 10t +2t =12t to obtain (t+1) 5

2*(t+1)³

The coefficient containing t in 1 is 12, i.e., a3=12

For the left x 5 coefficient is 1, x 3 coefficient is 2, x 4 coefficient is 0, so for the right side, x 5 coefficient is a5, so a5 = 1 according to the Yang Hui triangle, the coefficient of (x-1) 5, x 4 is -5, and the x3 coefficient is 10

Then the coefficient for a4(x-1) 4, x 4 is a4 and the coefficient for x3 is -6a4

The coefficient for a3(x-1) 3, x 3 is a3, so there is -5+a4=0;10-6a4+a3=0 can be calculated as a4=5, a3=20

Obviously a5=1, a4=0

So: a5(x-1) 5+a4(x-1) 4+a3(x-1) 3(x-1) 5+a 3(x-1) 3

The coefficient of x 3 = a3 + c(5,3) = a 3 + 10 so: a3 + 10 = 2

a3=-8

It's the third year of high school math, it's high, hehe, I just happen to split (2+x) 5 into (3+x-1) 5 again(3+x-1) 5=c5 take 0 3 5+c5 take 1 3 5·(x-1) 4++c5 takes 5 (x-1) 5 - Qi Shou Li ——— that mathematical symbol is not Qin Qin, you should understand it——— then a1=405a3=180a5=1

x has no limit, so let x=0 and the above equation becomes:

a0 a1(0 1) a2(0 1)^2 a3(0 1)^3 ..a2013(0 1)^2013=(0^2 0 1)^1006(0 2)

i.e. a0 a1 a2 a3 ...a2012 a2013=1^1006×2,5,tangram_guid_1360849629140???The assignment method is adopted.

Substituting x=0 gives the original formula =(0 2+0+1) 1006(0+2)=1,0, and lets a0+a1(x+1)+a2(x+1) 2+a3(x+1) 3+...a2013(x+1) 2013=(x 2+x+1) 1006(x+2), then a0+a1+a2+a3+.a2012=?

1-2x)^5=a0+a1(x-1)+a2(x-1)^2+.+a5(x-1) 5 then let f(x)=(1-2x) 5 find the derivative to get the trillion attack accompaniment: f'(x)=5[(1-2x) 4]*(2)=-10(1-2x) 4 and f'(x)=a1+2a2(x-1)+3a3(x-1) 2+4a4(x-1) 3+5a5(x-1) 4 makes the people stupid x=2 then:

1-2*2) 4=a1+2a2+3a3+4a4+5a5, i.e.: Zen early a1+2a....

To find the coefficients of x 3 on the right, simply add the coefficients of x 3 for each term on the left side of the equation, ie.

Analysis: From the meaning of the question to x=t+1, then:

t+1)^5 +2*(t+1)³ 1=a0+a1*t+a2*t² +a3*t³ +a4*t^4 +a5*t^5

Then it is easy to know that a3 is the coefficient containing t in (t+1) 5 +2*(t+1) 1.

The term containing t in binomial equation (t+1) 5 is c(5,2)*t =10t, and the term containing t in t+1) is t, that is, the term containing t in 2*(t+1) is 2t

Therefore, we can see that the term containing t in (t+1) 5 +2*(t+1) 1 is 10t +2t = 12t

The coefficient containing t in (t+1) 5 +2*(t+1) 1 is 12, that is, quietly cultivating a3=12

a5 = 1, to the fifth power gives -5x 4

There is no fourth-power term at the left end, so a4=5

This is the binomial part, and the way to do this is to identify a5 first

First of all, the left terms are (1+x), so the left side of the equation can be summed by the proportional sequence.

Using the summation formula of the proportional series, [(1+x) 51-(1+x) 3] x is obtained by simplification

Then use the binomial theorem. (1+x) The fourth power term of 51 is 1499400, (1+x) 3 does not have a fourth power term, and the fourth power term is a cubic term except for one square term, and the coefficient at this time is 1499400, so the value of A3 is 1499400 Thank you!