Solution: For (1+a) n, terms with equal coefficients satisfy "sum of terms = n+2" (there are n+ 1 terms in the expansion).
Therefore: 4r+ r+2=20+2.
Therefore: r=4.
2. Let m and n be positive integers, f (x) = (1-2x) m+(1-5x) n (where both m and n represent powers), and the coefficient of the first term containing x is-16. What is the coefficient of the x term?
Solution: The coefficient of the first term containing X in (1-2x) m is C[m]( 1)? (-2)=-2m
The coefficient of the first term containing x in (1-5x) n is C[n]( 1)? (-5)=-5n
Therefore: -2m-5n=- 16.
M and n are positive integers, so: n=2, m=3.
So: f (x) = (1-2x) m+(1-5x) n = (1-2x)? +( 1-5x)？
So the coefficient with the x term is C[3](2)? (-2)? +C[2](2)？ (-5)? =37
3. Known (x? -1√ 3x) n (n is a power), where the binomial coefficient of the fourth term is 220, then n=? What is the item that contains the x cube in the expansion?
Solution: (x? -1√ 3x) n (n stands for power), where the binomial coefficient of the fourth term is c [n] (3) = 220, so: n= 12.
So: (x? In the expansion of-1√ 3x)12, the term containing the cube of x is C[ 12](5)? (x？ )^5? (- 1/√3X) ^7=-88√3x？ /9
4.( 1-x)+( 1-x)？ +...+( 1-x) 10 power expansion, x? What is the coefficient of?
Solution: (1-x)+( 1-x)? +...+( 1-x) 10 power expansion, x? The coefficient is: c [2] (2)+c [3] (2)+c [4] (2)+…+c [10] (2) =1+3+6+10+/kloc.
You can also sum first and then find x? Coefficient of
5. It is known that the expansion of {a/x-√ (x/2)} 9th power contains x? The coefficient of the power term is 9/4, so what is the coefficient of the fifth term in the expansion?
Solution: Does the expansion of {a/x-√ (x/2)} 9th power contain X? Is the power term C[9]( 1)? (a/x)？ [-√(x/2))]^8=9ax？ / 16
Therefore: 9a/ 16=9/4, therefore: a=4.
So the fifth term in the {a/x-√ (x/2)} 9th power expansion is C[9](5)? (4/x) ^5？ [-√(x/2))]^4=32256/x？
So the coefficient of the fifth term in the expansion is 32256.
6. Find the polynomial (3x4-x3+2x? The eighth power of -3)? The 4th power of (3x-5)? The 6th power of (7x4-4x-2) is the sum of the coefficients in the expansion.
Solution: When x= 1, (3x4-x3+2x? The eighth power of -3)? The 4th power of (3x-5)? The 6th power value of (7x4-4x-2) is the sum of the coefficients in the expansion. That is, the eighth power of (3- 1+2-3)? The fourth power of (3-5)? The 6th power of (7-4-2) = 16.
7. Polynomial x 100 power -x+(-x3 power -2x? What is the sum of the coefficients of the even power of x and the odd power of x in the100th power expansion of +2?
Solution: Let the sum of even power coefficients of X be A, and the sum of odd power coefficients of X be B.
Then: when x= 1, a+b =11+(-13-2+2)100 =1.
When x=- 1, A-B = (-1)100+1[-(-1) 3-2+2]100 = 3.
Therefore: a=2, b=- 1.
That is, the sum of even power coefficients of X is 2, and the sum of odd power coefficients of X is-1.
8。 When n is a positive integer, it is proved that 2 ≤ [1+( 1/n)] n ≤ 3.
Prove: [1+(1/n)] n =1+c [n] (1)? ( 1/n)+C[n](2)？ ( 1/n) ^2+ C[n](3)？ ( 1/n) ^3+…+ C[n](n-2)？ ( 1/n)^(n-2)+ c[n](n- 1)？ ( 1/n) ^(n- 1)+ C[n](n)？ ( 1/n) ^n
=2+(n- 1)/(2n)+(n- 1)(n-2)/(6n^2)+….+(n- 1)/[2(n- 1)^(n-3)]+ 1/[n^(n-2)]+ 1/(n ^n)>2
On [1+( 1/n)] n ≤ 3 by mathematical induction.
9。 What are the last two digits of 99 100- 19?
Solution: 99100 = (100/)100 = ...+c [100] (2)? (65438+( 100)^ 1+c[ 100](0)
So: the last two digits of 99 100- 19 and C[ 100](2)? The last two digits of (65438+(100)1+c [100] (0)-19 are the same, that is, 82.
10。 What is the sum of all odd coefficients in the cubic power expansion of (1+x) +( 1+x)+(1+x)?
Refer to the seventh question
1 1。 It is known that n belongs to a positive integer. Prove: the n power of 2 >; 1+2+3+...+n。
By mathematical induction