Advertisements
Advertisements
प्रश्न
By the principle of mathematical induction, prove the following:
n(n + 1) (n + 2) is divisible by 6, for all n ∈ N.
Advertisements
उत्तर
P(n): n(n + 1) (n + 2) is divisible by 6.
P(1): 1 (2) (3) = 6 is divisible by 6
∴ P(1) is true.
Let us assume that P(k) is true for n = k
That is, k (k + 1) (k + 2) = 6m for some m
To prove P(k + 1) is true i.e. to prove (k + 1) (k + 2)(k + 3) is divisible by 6.
P(k + 1) = (k + 1) (k + 2) (k + 3)
= (k + 1)(k + 2)k + 3(k + 1)(k + 2)
= 6m + 3(k + 1)(k + 2)
In the second term either k + 1 or k + 2 will be even, whatever be the value of k.
Hence second term is also divisible by 6.
∴ P (k + 1) is also true whenever P(k) is true.
By Mathematical Induction P (n) is true for all values of n.
APPEARS IN
संबंधित प्रश्न
By the principle of mathematical induction, prove the following:
4 + 8 + 12 + ……. + 4n = 2n(n + 1), for all n ∈ N.
By the principle of mathematical induction, prove the following:
1 + 4 + 7 + ……. + (3n – 2) = `("n"(3"n" - 1))/2` for all n ∈ N.
By the principle of mathematical induction, prove the following:
32n – 1 is divisible by 8, for all n ∈ N.
By the principle of mathematical induction, prove the following:
52n – 1 is divisible by 24, for all n ∈ N.
By the principle of mathematical induction, prove the following:
2n > n, for all n ∈ N.
By the principle of mathematical induction, prove that, for n ≥ 1
12 + 32 + 52 + ... + (2n − 1)2 = `("n"(2"n" - 1)(2"n" + 1))/3`
Using the Mathematical induction, show that for any natural number n ≥ 2,
`(1 - 1/2^2)(1 - 1/3^2)(1 - 1/4^2) ... (1 - 1/"n"^2) = ("n" + 1)/2`
Using the Mathematical induction, show that for any natural number n ≥ 2,
`1/(1 + 2) + 1/(1 + 2 + 3) + 1/(1 +2 + 3 + 4) + .... + 1/(1 + 2 + 3 + ... + "n") = ("n" - 1)/("n" + 1)`
Using the Mathematical induction, show that for any natural number n,
`1/(2.5) + 1/(5.8) + 1/(8.11) + ... + 1/((3"n" - 1)(3"n" + 2)) = "n"/(6"n" + 4)`
Use induction to prove that 5n+1 + 4 × 6n when divided by 20 leaves a remainder 9, for all natural numbers n
