问题11。如果28 C 2r : 24 C 2r-4 = 225:11,则找到r。
解决方案:
(28!/(28-2r)!2r!)/(24!/(24-2r+4)!(2r-4)!)=225/11
28x27x26x25/2r(2r-1)(2r-2)(2r-3)=225/11
28x27x26x25x11=225x(2r)(2r-1)(2r-2)(2r-3)
28x3x26x11=2r(2r-1)(2r-2)(2r-3)
14x2x3x13x2x11=2r(2r-1)(2r-2)(2r-3)
14x13x12x11=2r(2r-1)(2r-2)(2r-3)
2r=14
r=7
问题12.如果AP中有n C 4 , n C 5和n C 6 ,则找到n。
解决方案:
We know that the A.P series is represented as a,a+d,a+2d,..
=>2.nC5=nC4+nC6
=>2(n!/(n-5)!5!)=(n!/(n-4)!4!)+n!/(n-6)!6!
=>2/(n-5)5=(1/(n-5)(n-4))+(1/30)
=>(2/(n-5)5)-(1/(n-5)(n-4))=1/30
=>(2(n-4)-5)30=(n-5)(n-4)5
=>60n-240-150=(n-5)(n-4)5
=>12n-78=(n-5)(n-4)
=>12n-78=n2-9n+20
=>n2-21n+98=0
=>n2-7n-14n+98=0
=>n(n-7)-14(n-7)=0
=>(n-7)(n-14)=0
=>n=7, 14
问题13。如果2n C 3 : n C 2 = 44:3,则找到n。
解决方案:
(2n!/(2n-3)!3!)/(n!/(n-2)!2!)=44/3
2n(2n-1)(2n-2)/3n(n-1)=44/3
2n(2n-1)(2n-2)=44n(n-1)
(2n-1)(2n-2)=22n-22
4n2-6n+2=22n-22
4n2-28n+24=0
n2-7n+6=0
(n-1)(n-6)=0
n=1 , 6
Let n=1
then 2(1)C3:2C2 is not possible because n So, n=6.
问题14.如果16 C r = 16 C r + 2 ,则找到r C 4 。
解决方案:
16=r+r+2
16=2r+2
14=2r
r=7
=>7C4=7!/3!4!
=5x6x7/3×2
=35
问题15.如果α= M C 2,然后找到α的C 2的值。
解决方案:
∝C2=∝!/(∝-2)!2!
=(∝-1)(∝)/2
=(mC2-1)(mC2)/2
=(m!/(m-1)!2!-1)(m!/(m-2)!2!)/2
=(m(m-1)-2)(m(m-1))/8
=(m2-m-2)(m(m-1))/8
=(m+1)(m-1)(m-2)m/8
=1/8[(m-2)(m-1)(m)(m+1)]
问题16.证明2n个连续负整数的乘积可被(2n)整除!
解决方案:
Let the 2n consecutive negative integers be -k,-k-1,-k-2,…,-k-(2n-1)
product of 2n consecutive negative integers=-kx-k-1x-k-2x….x-k-(2n-1)
=(-1)2nxkxk+1xk+2x…..xk+(2n-1)
=(-1)2nxkxk+1xk+2x…..xk+(2n-1)x(k-1)!/(k-1)!
=(-1)2nx(k+2n-1)!/(k-1)!
=(-1)2nx(k+2n-1)!(2n)!/(k-1)!(2n)!
=(-1)2nxk+2n-1C2nx(2n)!
Therefore the product of 2n consecutive negative integers is divided by (2n)!.
问题17。对于所有正整数n,证明2n C n + 2n C n-1 = 1/2 [ 2n + 2 C n + 1 ]。
解决方案:
LHS=2nCn+2nCn-1
=2n!/n!(2n-n)!+2n!/(n-1)!(2n-n+1)!
=2n!/n!n!+2n!/(n-1)!(n+1)!
=2n!/n(n-1)!n!+2n!/(n-1)!n!(n+1)
=2n!/n(n+1)(n-1)!n![n+1+n]
=2n!(2n+1)/n(n+1)(n-1)!n!
=(2n+1)!/n!(n+1)!
=(2n+2)(2n+1)!/n!(n+1)!(2n+2)
=(2n+2)!/n!(n+1)!(n+1)2
=(2n+2)!/(n+1)!(n+1)!2
=(2n+2)!/(n+1)!(2n+2-n-1)!2
=1/2[2n+2Cn+1]
= RHS
问题18.证明: 4n C 2n : 2n C n = [1x 3 x 5 x….. x 4n-1]:[1 x 3 x 5 x……x 2n-1] 2 。
解决方案:
LHS=4nC2n/2nCn
=(4n!/2n!2n!)/(2n!/n!n!)
=[1 x 2 x ……x 4n] x[1 x2x3x….xn]2/[1x2x…x2n]3
=[1x3x5x…4n-1][2x4x6x…4n](n!)(n!)/[1x3x5x…x2n-1]2[2x4x6x…x2n]2(2n!)
=[1x3x5x..x4n-1]22n[1x2x3x..2n](n!)(n!)/[1x3x5x…x2n-1]2x22n[1x2x3x..xn]2(2n)!
=[1x3x5x…x4n-1]/[1x3x5x2n-1]2
=RHS
问题19:评估
解决方案:
=>20C5+20C4+21C4+22C4+23C4
W.K.T nCr+nCr-1=n+1Cr
=>21C5+21C4+22C4+23C4
=>22C5+22C4+23C4
=>23C5+23C4
=> 24C5
问题20.1令r和n为正整数,使得1 <= r <= n。然后证明以下内容:
n C r / n C r-1 = n-r + 1 / r
解决方案:
LHS=(n!/r!(n-r)!)/(n!/(r-1)!(n-r+1)!
=(n-r)!(n-r+1)(r-1)!/(n-r)!(r)(r-1)!
=n-r+1/r
=RHS
问题20.2。令r和dn为正整数,使得1 <= r <= n。然后证明以下内容:
nx n-1 C r-1 =(n-r + 1) n C r-1
解决方案:
LHS=n(n-1)!/(r-1)!(n-1-r+1)!
=n!/(r-1)!( n-r)!
=n!(n-r+1)/(r-1)!( n-r+1)(n-r)!
=n!(n-r+1)/(r-1)!(n-r+1)!
=(n-r+1)nCr-1
=RHS
问题20.3令r和n为正整数,使得1 <= r <= n。然后证明以下内容:
n C r / n-1 C r-1 = n / r
解决方案:
LHS=(n!/r!(n-r)!)/(n-1)!/(n-r)!(r-1)!
=n(n-1)!(r-1)!(n-r)!/r(r-1)!(n-r)!(n-1)!
=n/r
=RHS
问题20.4。令r和n为正整数,使得1 <= r <= n。然后证明以下内容:
n C r +2 x n C r-1 + n C r-2 = n + 2 C r 。
解决方案:
W.K.T nCr+nCr-1=n+1Cr
LHS=nCr+nCr-1+nCr-1+nCr-2
=n+1Cr+n+1Cr-1
=n+2Cr
=RHS