问题1.计算：
(i)30！/ 28！
(ii)(11！– 10！)/ 9！
(iii)LCM(6 !, 7 !, 8！)
解决方案：
(i) 30!/28!
We know that,
n! = n(n-1)!
Therefore, 30! = 30 × 29! = 30 ×29×28!
30!/28! = (30 × 29 × 28!)/28!
= 30 × 29 = 870
(ii) (11! – 10!)/9!
We know that,
n! = n(n-1)!
Therefore,
11! = 11 × 10! =11 × 10 × 9!
10! = 10 × 9!
By using these values, we get
(11! – 10!)/9! = (11 × 10 × 9! – 10 × 9!)/ 9!
= 9! (110 – 10)/9!
= 110 – 10
= 100
(iii) L.C.M. (6!, 7!, 8!)
We know that,
8! = 8 × 7 × 6!
7! = 7 × 6!
6! = 6!
So,
L.C.M. of 6!, 7!, 8! = LCM [8 × 7 × 6!, 7 × 6!, 6!]
= 8 × 7 × 6!
= 8!
问题2：证明：1/9！ + 1/10！ + 1/11！ = 122/11！
解决方案：
Given:
1/9! + 1/10! + 1/11! = 122/11!
LHS: 1/9! + 1/10! + 1/11!
Using n = n(n-1)!
1/9! + 1/10! + 1/11! = 1/9! + 1/(10×9!) + 1/(11×10×9!)
= (110 + 11 + 1)/(11 × 10 × 9!)
= 122/11!
= RHS
Therefore, RHS = LHS
Hence, proved.
问题3.在以下每一个中找到x：
(i)1/4！ + 1/5！ = x / 6！
(ii)x / 10！ = 1/8！ + 1/9！
(iii)1/6！ + 1/7！ = x / 8！
解决方案：
(i) 1/4! + 1/5! = x/6!
We know that
5! = 5 × 4!
6! = 6 × 5!
So by using these values
1/4! + 1/5! = x/6!
1/4! + 1/(5×4!) = x/(6×5!)
(5 + 1) / (5×4!) = x/(6×5!)
6/5! = x/(6×5!)
x = (6 × 6 × 5!)/5!
= 36
∴ x = 36.
(ii) x/10! = 1/8! + 1/9!
We know that
10! = 10 × 9!
9! = 9 × 8!
Using these values, we get
x/10! = 1/8! + 1/9!
x/10! = 1/8! + 1/(9×8!)
x/10! = (9 + 1) / (9×8!)
x/10! = 10/9!
x/(10×9!) = 10/9!
x = (10 × 10 × 9!)/9!
= 10 × 10
= 100
∴ x = 100.
(iii) 1/6! + 1/7! = x/8!
We know that
8! = 8 × 7 × 6!
7! = 7 × 6!
So by using these values,
1/6! + 1/7! = x/8!
1/6! + 1/(7×6!) = x/8!
(1 + 7)/(7×6!) = x/8!
8/7! = x/8!
8/7! = x/(8×7!)
x = (8 × 8 × 7!)/7!
= 8 × 8
= 64
∴ The value of x is 64.
问题4.将以下乘积转换为阶乘：
(i)5⋅6⋅7⋅8⋅9⋅10
(ii)3⋅6⋅9⋅12⋅15⋅18
(iii)(n + 1)(n + 2)(n + 3)…(2n)
(iv)1⋅3⋅5⋅7⋅9…(2n – 1)
解决方案：
(i) 5 ⋅ 6 ⋅ 7 ⋅ 8 ⋅ 9 ⋅ 10
We can rewrite the above expression as:
5 ⋅ 6 ⋅ 7 ⋅ 8 ⋅ 9 ⋅ 10 = (1×2×3×4×5×6×7×8×9×10)/(1×2×3×4)
= 10!/4!
(ii) 3 ⋅ 6 ⋅ 9 ⋅ 12 ⋅ 15 ⋅ 18
We can rewrite the above expression as:
3 ⋅ 6 ⋅ 9 ⋅ 12 ⋅ 15 ⋅ 18 = (3×1) × (3×2) × (3×3) × (3×4) × (3×5) × (3×6)
= 36 (1×2×3×4×5×6)
= 36 (6!)
(iii) (n + 1) (n + 2) (n + 3) … (2n)
We can rewrite the above expression as:
(n + 1) (n + 2) (n + 3) … (2n) = [(1) (2) (3) ..(n) … (n + 1) (n + 2) (n + 3) … (2n)] / (1) (2) (3) .. (n)
= (2n)!/n!
(iv) 1 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 9 … (2n – 1)
We can rewrite the above expression as:
1 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 9 … (2n – 1) = [(1) (3) (5) … (2n-1)] [(2) (4) (6) … (2n)] / [(2) (4) (6) … (2n)]
Take 2 out from each term of denominator
= [(1) (2) (3) (4) … (2n-1) (2n)] / 2n [(1) (2) (3) … (n)]
= (2n)! / 2nn!
问题5.以下哪项是正确的：
(i)(2 + 3)！ = 2！ + 3！
(ii)(2×3)！ = 2！ ×3！
解决方案：
(i) (2 + 3)! = 2! + 3!
LHS:
(2 + 3)! = 5!
RHS,
2! + 3! = (2×1) + (3×2×1)
= 2 + 6
= 8
LHS ≠ RHS
∴ The given expression is false.
(ii) (2 × 3)! = 2! × 3!
LHS:
(2 × 3)! = 6!
= 6 × 5 × 4 × 3 × 2 × 1
= 720
RHS,
2! × 3! = (2×1) × (3×2×1)
= 12
LHS ≠ RHS
∴ The given expression is false.
问题6.证明：n！ (n + 2)= n！ +(n + 1)！
解决方案：
Given:
n! (n + 2) = n! + (n + 1)!
Method -1 (converting RHS to LHS)
RHS = n! + (n + 1)!
n! + (n + 1)! = n! + (n + 1) (n)!
= n!(1 + n + 1)
= n! (n + 2)
= L.H.S
L.H.S = R.H.S
Hence, Proved.
Method – 2 (converting LHS to RHS)
LHS = n! (n + 2)
n! (n + 2) = n! (1 + n + 1)
= n!(1)+n!(n + 1)
= n! +(n + 1)!
= R.H.S
R.H.S = L.H.S
Hence, Proved.
问题7：如果(n + 2)！ = 60 [(n – 1)！]，找到n。
解决方案：
We know that,
n! = n(n-1)!
By using this property
(n + 2)! = 60[(n – 1)!]
(n + 2)(n + 1)(n)[(n – 1)!] = 60[(n – 1)!]
(n + 2)(n + 1)(n) = 60
(n + 2)(n + 1)(n) = 5 × 4 × 3
On comparing both sides, we get
n = 3
∴ n = 3
问题8.如果(n + 1)！ = 90 [(n – 1)！]，找到n。
解决方案：
We know that,
n! = n(n-1)!
By using this property
(n + 1)! = 90[(n – 1)!]
(n + 1)(n)[(n – 1)!] = 90[(n – 1)!]
(n + 1)(n) = 90
(n + 1)(n) = 10 × 9
On comparing both sides,we get
n = 9
∴ n = 9
问题9.如果(n + 3)！ = 56 [(n + 1)！]，找到n。
解决方案：
We know that,
n! = n(n-1)!
By using this property
(n + 3)! = 56[(n + 1)!]
(n + 3)(n+2)[(n + 1)!] = 56[(n + 1)!]
(n + 3)(n + 2) = 56
(n + 3)(n + 2) = 8 × 7
On comparing both sides,we get
n + 2 = 7
∴ n = 5
问题10：如果(2n)！ /(3！(2n – 3)！)和n！ /(2！(n – 2)！)的比例为44：3，找到n。
解决方案：
Let [(2n)! / (3!(2n – 3)!)] / [n! / (2!(n – 2)!) ] = 44/3
[(2n)! × 2!(n – 2)!] / [3!(2n – 3)! × n!] = 44/3
[2n×(2n-1)×(2n-2)×(2n-3)!×2!×(n-2)!] / [3!×(2n-3)!×n×(n-1)×(n-2)!] = 44/3
(2n×(2n-1)×2(n-1))/(3×n×(n-1)) = 44/3
4(2n-1) = 44
2n-1 = 11
2n = 12
∴ n = 6
问题11：证明：
在！ /(nr)！ = n(n-1)(n-2)。 。 。 (n-(r-1))
ii)n！ /(((nr)！r！)+ n！ /(((n-r + 1)！(r-1)！)=(n + 1)！ /(r！(n-r + 1)！)
解决方案：
(i) n! / (n-r)! = n(n-1)(n-2) . . . (n-(r-1))
LHS:
n! / (n-r)! = [n×(n-1)×(n-2) . . . (n-r+2)×(n-r+1)×(n-r)!] / (n-r)!
= n×(n-1)×(n-2). . . (n-r+2)×(n-r+1)
= n×(n-1)×(n-2). . . (n-(r-2))×(n-(r-1))
= n(n-1)(n-2). . . (n-(r-1))
= RHS
LHS = RHS
Hence, proved.
(ii) n! / ((n-r)!r!) + n! / ((n-r+1)!(r-1)!) = (n+1)! / (r!(n-r+1)!)
LHS:
n! / ((n-r)!r!) + n! / ((n-r+1)!(r-1)!) = n![(n-r+1) / ((n-r+1)!r!) + r / ((n-r+1)!r!) ]
= n![(n-r+1+r) / (n-r+1)!r!]
= n!(n+!)/[(n-r+1)!r!]
= (n+1)! / (r!(n-r+1)!)
= RHS
LHS = RHS
Hence, proved.
问题12.证明：i)(2n + 1)！ / n！ = 2 n [1.3.5。 。 。 (2n – 1)(2n + 1)]
解决方案：
LHS:
(2n+1)! / n! = [1×2×3×4 . . . (2n-2)×(2n-1)×(2n)×(2n+1)] / n!
Separate even and odd terms in numerator
= {[2×4 ×6 . . . (2n-2)×(2n)][1×3 ×5 . . . (2n-1)×(2n+1)]} / n!
Take factor of two out from all even terms of numerator
= {2n [1×2×3. . . (n-1)×(n)] [1×3×5. . . (2n-1)×(2n+1)]} / n!
= {2n n! [1×3 ×5. . . (2n-1)×(2n+1)]}/n!
= 2n [1×3×5. . . (2n-1)×(2n+1)]
= RHS
LHS = RHS
Hence, proved.