(x+y)n =∑k = 0n \ math choice((((NK \ math choice))))xkyn？ k
Common forms
(x+ 1)n =∑k = 0n \ math choice((((NK \ math choice))))xk
Equivalent form
(x+y)n =∑k = 0n \ math choice((((NK \ math choice))))xkyn？ k=∑k=0n\mathchoice((((nn？ k\mathchoice))))xkyn？ k=∑k=0n\mathchoice((((nn？ k\mathchoice))))xn？ kyk =∑k = 0n \ math choice((((NK \ math choice)))xn？ kyk
Proof 1 (combined meaning/combined analysis/secondary calculation)
(x+y)n=(x+y)×(x+y)×？ ×(x+y)
For every xkyn? K, that is, choose k x from n (x+y), n? K y, so there is a \ math choice((NK \ math choice)) method, that is, there is \ mathchoice ((NK \ mathchoice)) xkyn? k
Certificate of completion
Proof 2 (Mathematical Induction)
When n= 1, the formula obviously holds.
Assuming that the formula is true for positive integer n, it is proved that the formula is also true for n+ 1
direct proofs
(x+y)n+ 1 =∑k = 0n+ 1 \ math choice((((NK \ math choice)))xkyn+ 1？ k
Because the formula applies to n, there are
(x+y)n(x+y)n+ 1 =∑k = 0n \ math choice((((NK \ math choice)))xkyn？ k =(x+y)∑k = 0n \ math choice((((NK \ math choice)))xkyn？ k = x∑k = 0n \ math choice((((NK \ math choice))))xkyn？ k+y∑k = 0n \ math choice((((NK \ math choice))))xkyn？ k =∑k = 0n \ math choice((((NK \ math choice)))xk+ 1yn？ k+∑k = 0n \ math choice((((NK \ math choice))))xkyn+ 1？ k =∑k = 0n \ math choice((((NK \ math choice)))xk+ 1y(n+ 1)？ (k+ 1)+∑k = 0n \ math choice((((NK \ math choice)))xkyn+ 1？ k =∑k+ 1 = 1n+ 1 \ math choice((((n(k+ 1)？ 1 \ math choice))))xk+ 1y(n+ 1)？ (k+ 1)+∑k = 0n \ math choice((((NK \ math choice)))xkyn+ 1？ k =∑k = 1n+ 1 \ math choice((((NK？ 1 \ math choice))))xkyn+ 1？ k+∑k = 0n \ math choice((((NK \ math choice))))xkyn+ 1？ k=∑k= 1n\mathchoice((((nk？ 1 \ math choice))))xkyn+ 1？ k+xn+ 1+∑k = 1n \ math choice((((NK \ math choice)))xkyn+ 1？ k+yn+ 1 = \ math choice((((n+ 10 \ math choice)))xn+ 1+∑k = 1n \ math choice((((NK？ 1 \ math choice))))xkyn+ 1？ k+∑k = 1n \ math choice((((NK \ math choice)))xkyn+ 1？ k+\ math choice((((n+ 1n+ 1 \ math choice))))yn+ 1
By Pascal formula \ mathchoice ((n+1k \ mathchoice)) = \ mathchoice ((NK? 1\ mathchoice))+\ mathchoice (((NK \ mathchoice))) (which will be proved later), items with equal k can be merged.
(x+y)n+ 1 = \ math choice((((n+ 10 \ math choice)))xn+ 1+∑k = 1n \ math choice((((NK？ 1 \ math choice))))xkyn+ 1？ k+∑k = 1n \ math choice((((NK \ math choice)))xkyn+ 1？ k+\ math choice((((n+ 1n+ 1 \ math choice)))yn+ 1 = \ math choice(((n+ 10 \ math choice)))xn+ 1 y0+∑k = 1n \ math choice((((n+ 1k \ math choice)))xkyn+ 1k+\ math choice((((n+ 1n+ 1 \ math choice))))xyn+ 1 =∑k = 0n+ 1 \ math choice((((n+ 1k \ math choice))))xkyn+ 1？ k
Certificate of completion
Combinational identity
Formula 1
\ math choice((((NK \ math choice)))= \ math choice((((nn？ k\mathchoice))))
Proof (combined meaning)
Choosing K ball from N is equivalent to choosing N? K, don't stay
Certificate of completion
Formula 2
\ math choice((((NK \ math choice)))= NK \ math choice((((n？ 1k？ 1\mathchoice))))
Proof (formula method)
\ math choice((((NK \ math choice)))= n！ (n？ k)！ ×k！ =nk×(n？ 1)! (n？ k)！ ×k！ =nk×(n？ 1)! [(n？ 1)? (k？ 1)]! ×k！ =nk\mathchoice((((n？ 1k？ 1\mathchoice))))
Certificate of completion
Formula 3 (Pascal formula)
\ math choice((((NK \ math choice)))= \ math choice((((n？ 1k \ math choice)))+\ math choice((((n？ 1k？ 1\mathchoice))))
Proof (combined meaning)
Choosing k from n balls is equivalent to (not choosing the last one, the first n? 1 choose k) and (choose the last one, the nth one? 1 choose k? Union of 1
Formula 4
∑k = 0n \ math choice((((NK \ math choice)))= 2n
Proof (formula method)
(x+y) n = ∑ k = 0n \ mathchoice ((NK \ mathchoice)) xkyn from binomial theorem? K, let x=y= 1, then the equation becomes
( 1+ 1)n2n∑k = 0n \ math choice((((NK \ math choice)))=∑k = 0n \ math choice((((NK \ math choice)))× 1k× 1n？ k =∑k = 0n \ math choice((((NK \ math choice))))= 2n
Certificate of completion
Formula 5
∑k=0n(？ 1)k \ math choice((((NK \ math choice)))= 0
Proof (formula method)
(x+y) n = ∑ k = 0n \ mathchoice ((NK \ mathchoice)) xkyn from binomial theorem? K, let x=? 1, y= 1, then the equation becomes
[(? 1)+ 1]n0n∑k=0n(？ 1)k \ math choice((((NK \ math choice)))=∑k = 0n \ math choice((((NK \ math choice)))×(？ 1)k× 1n？ k=∑k=0n(？ 1)k \ math choice((((NK \ math choice)))= 0
Certificate of completion
Equation 6 (Sum of the following)
∑k = 0nk \ math choice((((NK \ math choice)))= n2n？ 1
Proof 1 (combined meaning)
Formula meaning: the sum of the times that each ball appears in all the selection methods of selecting n balls.
Correct meaning: how many times do you choose n balls and 2n balls? 1
Left-handed meaning: for the case of choosing K ball in all options, its contribution to the answer is K, * * * has \ math choice((NK \ math choice)).
Obviously, the three meanings are equivalent.
Certificate of completion
Proof 2 (Formula Method)
By formula 2 \ mathchoice ((NK \ mathchoice)) = NK \ mathchoice ((n? 1k？ 1\mathchoice))):
∑k = 0nk \ math choice(((NK \ math choice)))= 0 \ math choice(((NK \ math choice)))+∑k = 1nk×\ math choice(((NK \ math choice)))= 0+∑k = 1nk×NK \ math choice(((n？ 1k？ 1 \ math choice)))=∑k = 1nn \ math choice((((n？ 1k？ 1 \ math choice))))= n∑k = 1n \ math choice((((n？ 1k？ 1\mathchoice))))=n∑k=0n？ 1\mathchoice((((n？ 1k\mathchoice))))
∑k=0n from formula 4 \ mathchoice ((NK \ mathchoice)) = 2n? 1\mathchoice((((n？ 1k\mathchoice))))=2n？ 1, so
∑k = 0nk \ math choice((((NK \ math choice)))= n∑k = 0n？ 1\mathchoice((((n？ 1k\mathchoice))))=n2n？ 1
Certificate of completion
Formula 7
∑k = 0nk 2 \ math choice((((NK \ math choice)))= n(n+ 1)2n？ 2
Proof 1 (combined meaning)
n(n+ 1)2n？ 2=n(n？ 1)2n？ 2+n×2×2n？ 2=n(n？ 1)2n？ 2+n×2n？ 1
Left-handed meaning: there are n different numbers. When k different numbers are selected, k2 ordered number pairs can be formed. The contribution of each ordered number pair (a, b) to the answer is 1, and the total contribution to the answer is k2. There are \ math choice((NK \ math choice)) ways to choose k different numbers.
Correct meaning: consider the contribution of each ordered number pair (a, b) to the answer; If a≠b, for other n? In each selection method of two numbers, there are 1 contributions, 2n? Two selection methods; If a=b, for other n? In each selection method of two numbers, there are 1 contributions, 2n? 1 selection method
The union of two cases of right meaning is equal to that of left meaning, so the two meanings are equivalent.
Certificate of completion
Proof 2 (Formula Method)
By formula 2 \ mathchoice ((NK \ mathchoice)) = NK \ mathchoice ((n? 1k？ 1\mathchoice))):
∑k = 0nk 2 \ math choice((((NK \ math choice)))= 0+∑k = 1nk 2×NK \ math choice((((n？ 1k？ 1 \ math choice)))=∑k = 1nk 2×NK \ math choice((((n？ 1k？ 1 \ math choice)))=∑k = 1nk×n \ math choice((((n？ 1k？ 1 \ math choice))))= n∑k = 1nk \ math choice((((n？ 1k？ 1\mathchoice))))=n∑k=0n？ 1(k+ 1)\ math choice((((n？ 1k\mathchoice))))=n∑k=0n？ 1k\mathchoice((((n？ 1k \ math choice)))+n∑k = 0n？ 1\mathchoice((((n？ 1k\mathchoice))))
By formula 6 ∑ k = 0nk \ mathchoice ((NK \ mathchoice)) = n2n? 1: ∑k=0n？ 1k\mathchoice((((n？ 1k\mathchoice))))=(n？ 1)2n？ 2
From formula 4 ∑ k = 0n \ Mathchoice ((NK \ Mathchoice)) = 2n: ∑ k = 0n? 1\mathchoice((((n？ 1k\mathchoice))))=2n？ 1
Therefore, the original formula can be changed to:
∑k = 0nk 2 \ math choice((((NK \ math choice)))= n∑k = 0n？ 1k\mathchoice((((n？ 1k \ math choice)))+n∑k = 0n？ 1\mathchoice((((n？ 1k\mathchoice))))=n(n？ 1)2n？ 2+n2n？ 1=n(n？ 1)2n？ 2+n×2×2n？ 2=n(n+ 1)2n？ 2
Certificate of completion
Equation 8 (Sum of Variable Terms)
∑l = 0n \ math choice((((lk \ math choice)))= \ math choice((((N+ 1k+ 1 \ math choice)))N，k∈N
Proof (combined meaning)
Choose k+ 1 ball from n+ 1 ball.
For all the selection methods, when the position of the last ball is considered as l+ 1(0≤l≤n), the contribution to the answer is the number of options for choosing K in the first L balls.
Therefore, for the union of all sets to which L belongs, it is equivalent to selecting the set of k+ 1 from n+ 1 balls.
Certificate of completion
Formula 9
\ math choice((((NR \ math choice)))\ math choice((((rk \ math choice)))= \ math choice((((NK \ math choice)))\ math choice((((n？ kr？ k\mathchoice))))
Proof (combined meaning)
Divide n balls into three piles, so that 1 pile has k balls, and the second pile has r? K ball, n in the third pile? Scheme number of r ball
Left-handed meaning: first choose r balls from these n balls, and put the remaining n? R is assigned to the third heap; Then choose k from R to 1 heap, and the rest R? K is assigned to the second heap.
Right-handed meaning: firstly, select K balls from N balls and assign them to 1 heap; And then from the rest of the n? Choose r from k? K is assigned to the second pile, and the remaining n? R is assigned to the third heap.
Obviously, the two meanings are equivalent without going into details.
Certificate of completion
Formula 10
∑k = 0r \ math choice((((MK \ math choice)))\ math choice((((NR？ k \ math choice)))= \ math choice((((m+NR \ math choice))))
Proof (combined meaning)
Correct meaning: choose r balls from m+n balls.
Left meaning: choose k balls from the first m balls, and choose r from the last n balls? The union of the union set of k balls (the first union set is for the number of schemes of each k, and the second union set is for ∑).
Obviously, these two meanings are equivalent.
Certificate of completion
Formula 1 1
∑k = 0m \ math choice((((MK \ math choice)))\ math choice(((NK \ math choice)))= \ math choice((((m+nm \ math choice))))
Proof (formula method)
∑k = 0m \ math choice((((MK \ math choice)))\ math choice((((NK \ math choice)))=∑k = 0m \ math choice((((mm？ k \ math choice)))\ math choice((((NK \ math choice))))
By the formula10 ∑ k = 0r \ mathchoice ((MK \ mathchoice)) \ mathchoice ((NR? k \ math choice)))= \ math choice((m+NR \ math choice)))∑k = 0r \ math choice((Mr？ K \ mathchoice))) \ mathchoice ((NK \ mathchoice)) = \ mathchoice ((n+Mr \ mathchoice)), let r=m: ∑ k = 0m \ mathchoice (()). K \ mathchoice))) \ mathchoice ((NK \ mathchoice)) = \ mathchoice ((m+nm \ mathchoice))), so.
∑k = 0m \ math choice((((MK \ math choice)))\ math choice((((NK \ math choice)))=∑k = 0m \ math choice((((mm？ k \ math choice)))\ math choice((((NK \ math choice)))= \ math choice((((m+nm \ math choice))))
Certificate of completion
homework
Exercise 1
(3x？ What is the coefficient of x5y 13 in the expansion of 2y) 18? What is the coefficient of x8y9?
answer
Solution:
The coefficient of x5y 13 is \ mathchoice ((185 \ mathchoice)) (35+213), and the coefficient of x8y9 is 0.
Exercise 2
Proved by binomial theorem: 3n = ∑ k = 0n \ mathchoice ((NK \ mathchoice)) 2k.
Find ∑ k = 0n \ mathchoice ((NK \ mathchoice)) rk for any real number r.
answer
Certificate:
Commonly used binomial theorems: (x+1) n = ∑ k = 0n \ mathchoice ((NK \ mathchoice)) xk.
Let x=2, then the equation becomes (2+1) n = ∑ k = 0n \ mathchoice ((NK \ mathchoice)) 2k, that is, 3n = ∑ k = 0n \ mathchoice ((NK \ mathchoice)).
Certificate of completion
Solution:
Binomial theorems are commonly used: ∑ k = 0n \ mathchoice ((NK \ mathchoice)) rk = (r+1) n.
Exercise 3
Proved by binomial theorem: 2n=∑k=0n (? 1)k \ math choice((((NK \ math choice)))3n？ k
answer
Certificate:
From binomial theorem: (x+y) n = ∑ k = 0n \ Mathchoice ((NK \ Mathchoice)) xkyn? k
Let x=? 1, y=3, the equation becomes [(? 1)+3]n =∑k = 0n \ math choice((((NK \ math choice)))(？ 1)k3n？ k
That is 2n=∑k=0n (? 1)k \ math choice((((NK \ math choice)))3n？ k
Certificate of completion
Exercise 4
Find ∑k= 1n (? 1)k \ math choice((((NK \ math choice)))) 10k
answer
Solution:
∑k=0n(？ 1)k \ math choice((((NK \ math choice)))) 10k =∑k = 0n \ math choice((((NK \ math choice)))(？ 10)k
Commonly used binomial theorems: (x+1) n = ∑ k = 0n \ mathchoice ((NK \ mathchoice)) xk.
Let x=? 10, the equation becomes [(? 10)+ 1]n =∑k = 0n \ math choice((((NK \ math choice)))(？ 10)k
So ∑ k = 0n \ Mathchoice ((NK \ Mathchoice)) (? 10)k=(？ 9) Nouns
Exercise 5
Prove \ math choice((NK \ math choice)) with combinatorial meaning? \mathchoice((((n？ 3k \ math choice))))= \ math choice((((n？ 1k？ 1 \ math choice)))+\ math choice((((n？ 2k？ 1 \ math choice)))+\ math choice((((n？ 3k？ 1\mathchoice))))
answer
Certificate:
Choose k balls from n balls.
Left meaning: (total number of schemes)? (The number of schemes in which the first three balls are not selected), that is, the number of schemes in which at least one of the first three balls is selected.
Right meaning: (choose the scheme number of ball 1)+(choose the scheme number of ball 2 instead of ball 1)+(choose the scheme number of ball 3 instead of ball 1 and ball 2).
Obviously, these two meanings are equivalent.
Certificate of completion
Exercise 6
Let n be a positive integer, please prove:
∑k=0n(？ 1)k \ math choice((((NK \ math choice))))2 =？ 0，n=2m+ 1，m∈N+(？ 1)m \ math choice((((2mm \ math choice)))，n=2m，m∈N+
answer
Certificate:
If n is odd, then k and n? K is different parity.
∴ (? 1)k and (? 1)n？ K is one positive and one negative.
∑k=0n(？ 1)k(nk)2=∑k=0m(？ 1)k(nk)+∑k=m+ 1n(？ 1)k(nk)=∑k=0m(？ 1)k(nk)？ ∑k=m+ 1n(？ 1)n？ k(nn？ k)=∑k=0m(？ 1)k(nk)？ ∑k=0m(？ 1)k(nk)=0
If n is an even number, then k and n? K is the same parity.
∴ (? 1)k=(？ 1)n？ k
According to binomial theorem:
(x+ 1)n(x？ 1)n(x2？ 1)n=∑k=0n(nk)xk=∑k=0n(？ 1)n？ k(nk)xk=∑k=0n(？ 1)k(nk)xk=∑k=0n(？ 1)n？ k(nk)(x2)k=∑k=0n(？ 1)k(nk)(x2)k=∑k=0n(？ 1)k(nk)x2k
Therefore, the equation (x+ 1)n(x? 1)n=(x2？ 1)n
(x+ 1)n(x？ 1)n∑k=0n(nk)xk×∑k=0n(？ 1)k(nk)xk=(x2？ 1)n=∑k=0n(？ 1)k(nk)x2k
The left polynomial and the right polynomial are equal.
∴? The coefficients of I and xi are equal.
Let i=n=2m, that is, consider the coefficient of xn.
∑k=0n(nn？ k)xn？ k×(? 1)k(nk)xk∑k=0n(？ 1)k(nk)xn？ k×(nk)xk∑k=0n(？ 1)k(nk)2xn∑k=0n(？ 1)k(nk)2xn=(？ 1)m(nm)xn=(？ 1)m(nm)xn=(？ 1)m(nm)xn=(？ 1) m (2mm) xn
Go to xn at the same time and get
∑k=0n(？ 1)k(nk)2=(？ 1) m (2mm)
Certificate of completion
Exercise 7
Simplify (nk)+3(nk? 1)+3(nk？ 2)+(nk？ 3)
answer
Certificate:
(nk)+3(nk？ 1)+3(nk？ 2)+(nk？ 3)=(30)(nk)+(3 1)(nk？ 1)+(32)(nk？ 2)+(33)(nk？ 3)=
There are 2 piles of balls, 1 pile 3, and the second pile n.
Choosing k balls from n+3 is equivalent to (1 0 in the heap, k in the second heap)+(1 in the heap1,k in the second heap? 1)+( 1 Choose 2 in the heap, and choose K in the second heap? 2)+( 1 Choose 3 in the heap, and choose K in the second heap? three