Then we get (2×12-3×1+1) 3 = A0×16+a1×15+A2×14.
=a0+a 1+a2+a3+a4+a5+a6=0？ ①
Let x=- 1
Similarly, A0-a1+A2-A3+A4-A5+A6 = 216②.
①-② Get 2(a 1+a3+a5)=-2 16.
a 1+a3+a5=- 108
∴ Answer: The value of a 1+a3+a5 is-108.
2. Write the column number above.
1,2? Two numbers
1,2,3,2,? Four numbers
1, 2,3,4,3,2,6 numbers
1,2,3,4,5,4,3,2,? Eight numbers
1,2,3,4,5,6,5,4,3,2,? 10 number
1,2,3,....n......,3,2,? (n- 1)×2 numbers
The total * * * is 2+4+6+8+10+...+(n-1) × 2 = 2+(n-1)× 2 × (n-1).
While 20 10 is between 45×(45- 1)= 1980 and 46×(46- 1)=2070.
Therefore, the number 20 10 is in the column 1, 2, 3, 4, ....................................................................................................................................
∫20 10- 1980 = 30 (note: if the calculated number is greater than 46, the next number will be reduced).
The number 20 10 is the 30th number in this column.
The number reported by students on 20 10 is 30.
There seem to be fewer problems. With the first two conditions, it can be solved, and the last condition is useless.
Because the length of the iron wire is a and the three sides of the equilateral triangle are equal, the side length of the equilateral triangle is a/3.
As shown in the figure, △ABC is an equilateral triangle, and A is AD⊥BC in D.
Because an equilateral triangle has three lines on one line.
∴AD is also the center line of BC,
∴BD=CD= 1/2？ ×BC=a/6
∫AB = a/3
∴ AD & From Pythagorean Theorem; sup2+BD & amp； sup2AB & ampsup2
Ad = root symbol (ab & amp； sup2-BD & amp； sup2)
AD > 0
∴AD= root symbol (ab & amp； sup2-BD & amp； Sup2)=(a/2 root number 3= root number 3? ×a)/2
∫ s =1/2× ad× BC =1/2× (root 3a/2)×? a/3？ =9
∴ Solution A = 6× 3 (1/4)
∴a:a: The value of a is 6× 3 (1/4).
4. The problem should lack one condition: A and N are positive integers, otherwise the problem will not be solved.
∫ An n- 1 is a prime number.
There are two situations,
(1) an n- 1 is even, that is, 2.
Then n = 3,
And 3 are not divisible by an n such that n > 1, a > 1 and both a and n are positive integers.
So this situation is impossible.
② A n n-1is odd, then An n is even, so A must be even.
∫ can be obtained from the n-degree variance formula.
a^n- 1=[a^(n- 1)× 1^0+a^(n-2)× 1^ 1+a^(n-3)× 1^2+…+a^0× 1^(n- 1)](a- 1)
=[a^(n- 1)+a^(n-2)+a^(n-3)+…+a^0](a- 1)
On the other hand, there is = [a 0+a1+a 2+...+a (n-1)] (a-1).
∫ An n- 1 is a prime number.
It can only be decomposed into 1× a certain number.
∴ a (n-1)+a (n-2)+a (n-3)+…+a0 =1or a- 1= 1.
Obviously, a (n-1)+a (n-2)+a (n-3)+…+A0 =1is impossible.
∴ There must be-1= 1
∴a=2
Now, if n is a composite number, then n = xy, x and y are all positive integers greater than 1.
Then n- 1 is rewritten as
(a x) y- 1 = (a x- 1) [... polynomial]
∵ x > 1, where x is a positive integer and a=2.
∴ a x- 1 must be greater than 2.
Then the polynomial must also be greater than 2.
Then (a x) y- 1 can be decomposed into the product of several monomials > 2.
Then (a x) y- 1 is not a prime number, but a composite number.
∴n inseparable xy, x and y are all positive integers greater than 1
N is a prime number
To sum up: a=2, n must be a prime number.
5. connect FD, let AF=x, CF=y,
E is the midpoint of AD.
∴AE=DE
∴S△ABE=S△BDE,S△AEF=S△DEF
∴S shadow = s △ AEF+s △ BDE = s △ AEF+s △ Abe = s △ ABF = x/(x+y)
And s shadow = s △ AEF+s △ BDE = s △ BDE+s △ def = s △ BDF = y/(x+y)? ×a/(a+b)=ay/(x+y)(a+b)
∴ get:
xay
- === -
x+y？ (x+y)(a+b)
ax+bx=ay
Solution: y = (a+b) x/a+b) x/a.
And s shadow = x/(x+y) = x/[x+(a+b) x/a] = x/[(2a+b) x/a] = a/(2a+b)