Solution: It is given that: n! (n + 2) = n! + (n + 1)! Let us take the RHS first and solve accordingly! $ n!\text{ }+\text{ }\left( n\text{ }+\text{ }1 \right)!\text{ }=\text{ }n!\text{ }+\text{...
Prove that: n! (n + 2) = n! + (n + 1)!
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Exercise 16.1
Which of the following are true: (i) (2 + 3)! = 2! + 3! (ii) (2 × 3)! = 2! × 3!
Solution: (i) (2 + 3)! = 2! + 3! Let us consider LHS: (2 + 3)! (2 + 3)! = 5! Now RHS, 2! + 3! = (2×1) + (3×2×1) = 2 + 6 = 8 LHS ≠ RHS ∴ The given expression is false. (ii) (2 × 3)! = 2! × 3! Let us...
Convert the following products into factorials: (iii) (n + 1) (n + 2) (n + 3) …(2n) (iv) 1 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 9 … (2n – 1)
Solution: (iii) (n + 1) (n + 2) (n + 3) … (2n) We have: $ \left( n\text{ }+\text{ }1 \right)\text{ }\left( n\text{ }+\text{ }2 \right)\text{ }\left( n+3 \right)\ldots \left( 2n \right)=\left[ \left(...
Convert the following products into factorials: (i) 5 ⋅ 6 ⋅ 7 ⋅ 8 ⋅ 9 ⋅ 10 (ii) 3 ⋅ 6 ⋅ 9 ⋅ 12 ⋅ 15 ⋅ 18
Solution: (i) 5 ⋅ 6 ⋅ 7 ⋅ 8 ⋅ 9 ⋅ 10 We can write the above equation as: $ 5~\cdot ~6~\cdot ~7~\cdot ~8~\cdot ~9~\cdot ~10=\left( 1\times 2\times 3\times 4\times 5\times 6\times 7\times 8\times...
Find x in each of the following: (iii) 1/6! + 1/7! = x/8!
Solution: We can write that: 8! = 8 × 7 × 6! 7! = 7 × 6! So by making use of these values, $ 1/6!\text{ }+\text{ }1/7!\text{ }=\text{ }x/8! $ $ 1/6!\text{ }+\text{ }1/\left( 7\times 6! \right)\text{...
Find x in each of the following: (i) 1/4! + 1/5! = x/6! (ii) x/10! = 1/8! + 1/9!
Solution: (i) 1/4! + 1/5! = x/6! We can write 4! and 5! as: 5! = 5 × 4 × 3 × 2 × 1 6! = 6 × 5 × 4 × 3 × 2 × 1 So by making use of these values, $ 1/4!\text{ }+\text{ }1/5!\text{ }=\text{ }x/6! $ $...
Prove that: 1/9! + 1/10! + 1/11! = 122/11!
Solution: We are given that: 1/9! + 1/10! + 1/11! = 122/11! First, let us consider the LHS: $ 1/9!\text{ }+\text{ }1/10!\text{ }+\text{ }1/11!\text{ }=\text{ }1/9!\text{ }+\text{ }1/\left( 10\times...
Compute: (iii) L.C.M. (6!, 7!, 8!)
Solution: We have to find the LCM of (6!, 7!, 8!) We know that we can write: 8! = 8 × 7 × 6! 7! = 7 × 6! So, making use of the above values we get: $ L.C.M.\text{ }\left( 6!,\text{ }7!,\text{ }8!...
Compute: (i) 30!/28! (ii) (11! – 10!)/9!
Solution: (i) 30!/28! Using the properties of factorials! $ 30!/28!=\left( 30\times 29\times 28! \right)/28! $ $ 30!/28!=30\times 29 $ $ 30!/28!=870 $ (ii) (11! – 10!)/9! Using the properties of...