Advertisements
Advertisements
प्रश्न
n(n + 1) (n + 5) is a multiple of 3 for all n ∈ N.
Advertisements
उत्तर
Let P(n) be the given statement.
Now,
\[P(n): n(n + 1)(n + 5)\text{ is a multiple of } 3 . \]
\[\text{ Step1 } : \]
\[P(1): 1(1 + 1)(1 + 5) = 12 \]
\[\text{ It is a multiple of} 3 . \]
\[\text{ Hence, P(1) is true } . \]
\[\text{ Step2 } : \]
\[\text{ Let } P\left( m \right) \text{ be true . } \]
\[\text{ Then,} m\left( m + 1 \right)\left( m + 5 \right) \text{ is a multiple of } 3 . \]
\[\text{ Suppose} m\left( m + 1 \right)\left( m + 5 \right) = 3\lambda, \text{ where } \lambda \in N . \]
\[\text{ We have to show that } P\left( m + 1 \right) \text{ is true whenever P(m) is true } . \]
\[\text{ Now, } \]
\[P(m + 1) = \left( m + 1 \right)\left( m + 2 \right)\left( m + 6 \right)\]
\[ = m\left( m + 1 \right)\left( m + 6 \right) + 2\left( m + 1 \right)\left( m + 6 \right)\]
\[ = m\left( m + 1 \right)\left( m + 5 + 1 \right) + 2\left( m + 1 \right)\left( m + 6 \right)\]
\[ = m\left( m + 1 \right)\left( m + 5 \right) + m(m + 1) + 2\left( m + 1 \right)\left( m + 6 \right)\]
\[ = 3\lambda + \left( m + 1 \right)\left( m + 2m + 6 \right) \left[ From P(m) \right]\]
\[ = 3\lambda + 3\left( m + 1 \right)\left( m + 2 \right)\]
\[\text{ It is clearly a multiple of 3} . \]
\[\text{ Thus, P(m + 1) is true } . \]
\[\text{ By the principle of mathematical induction, P(n) is true for all n } \in N .\]
APPEARS IN
संबंधित प्रश्न
Prove the following by using the principle of mathematical induction for all n ∈ N: 1.2.3 + 2.3.4 + … + n(n + 1) (n + 2) = `(n(n+1)(n+2)(n+3))/(4(n+3))`
Prove the following by using the principle of mathematical induction for all n ∈ N: 1.2 + 2.22 + 3.22 + … + n.2n = (n – 1) 2n+1 + 2
Prove the following by using the principle of mathematical induction for all n ∈ N: `1/2 + 1/4 + 1/8 + ... + 1/2^n = 1 - 1/2^n`
Prove the following by using the principle of mathematical induction for all n ∈ N:
Prove the following by using the principle of mathematical induction for all n ∈ N:
Prove the following by using the principle of mathematical induction for all n ∈ N: 32n + 2 – 8n– 9 is divisible by 8.
If P (n) is the statement "n2 + n is even", and if P (r) is true, then P (r + 1) is true.
12 + 22 + 32 + ... + n2 =\[\frac{n(n + 1)(2n + 1)}{6}\] .
1 + 3 + 5 + ... + (2n − 1) = n2 i.e., the sum of first n odd natural numbers is n2.
\[\frac{1}{1 . 4} + \frac{1}{4 . 7} + \frac{1}{7 . 10} + . . . + \frac{1}{(3n - 2)(3n + 1)} = \frac{n}{3n + 1}\]
\[\frac{1}{3 . 7} + \frac{1}{7 . 11} + \frac{1}{11 . 5} + . . . + \frac{1}{(4n - 1)(4n + 3)} = \frac{n}{3(4n + 3)}\]
2 + 5 + 8 + 11 + ... + (3n − 1) = \[\frac{1}{2}n(3n + 1)\]
\[\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + . . . + \frac{1}{2^n} = 1 - \frac{1}{2^n}\]
a + (a + d) + (a + 2d) + ... (a + (n − 1) d) = \[\frac{n}{2}\left[ 2a + (n - 1)d \right]\]
2.7n + 3.5n − 5 is divisible by 24 for all n ∈ N.
11n+2 + 122n+1 is divisible by 133 for all n ∈ N.
Prove that n3 - 7n + 3 is divisible by 3 for all n \[\in\] N .
7 + 77 + 777 + ... + 777 \[{. . . . . . . . . . .}_{n - \text{ digits } } 7 = \frac{7}{81}( {10}^{n + 1} - 9n - 10)\]
x2n−1 + y2n−1 is divisible by x + y for all n ∈ N.
Prove by method of induction, for all n ∈ N:
3 + 7 + 11 + ..... + to n terms = n(2n+1)
Prove by method of induction, for all n ∈ N:
13 + 33 + 53 + .... to n terms = n2(2n2 − 1)
Prove by method of induction, for all n ∈ N:
(24n−1) is divisible by 15
Prove by method of induction, for all n ∈ N:
3n − 2n − 1 is divisible by 4
Prove by method of induction, for all n ∈ N:
Given that tn+1 = 5tn + 4, t1 = 4, prove that tn = 5n − 1
Answer the following:
Prove, by method of induction, for all n ∈ N
12 + 42 + 72 + ... + (3n − 2)2 = `"n"/2 (6"n"^2 - 3"n" - 1)`
Answer the following:
Prove by method of induction loga xn = n logax, x > 0, n ∈ N
Prove statement by using the Principle of Mathematical Induction for all n ∈ N, that:
1 + 3 + 5 + ... + (2n – 1) = n2
Prove the statement by using the Principle of Mathematical Induction:
23n – 1 is divisible by 7, for all natural numbers n.
Prove the statement by using the Principle of Mathematical Induction:
n2 < 2n for all natural numbers n ≥ 5.
Prove the statement by using the Principle of Mathematical Induction:
`sqrt(n) < 1/sqrt(1) + 1/sqrt(2) + ... + 1/sqrt(n)`, for all natural numbers n ≥ 2.
A sequence b0, b1, b2 ... is defined by letting b0 = 5 and bk = 4 + bk – 1 for all natural numbers k. Show that bn = 5 + 4n for all natural number n using mathematical induction.
Prove that for all n ∈ N.
cos α + cos(α + β) + cos(α + 2β) + ... + cos(α + (n – 1)β) = `(cos(alpha + ((n - 1)/2)beta)sin((nbeta)/2))/(sin beta/2)`.
If 10n + 3.4n+2 + k is divisible by 9 for all n ∈ N, then the least positive integral value of k is ______.
For all n ∈ N, 3.52n+1 + 23n+1 is divisible by ______.
