Advertisements
Advertisements
प्रश्न
Prove the following by using the principle of mathematical induction for all n ∈ N (2n +7) < (n + 3)2
Advertisements
उत्तर
Let the given statement be P(n), i.e.,
P(n): (2n +7) < (n + 3)2
It can be observed that P(n) is true for n = 1 since 2.1 + 7 = 9 < (1 + 3)2 = 16, which is true.
Let P(k) be true for some positive integer k, i.e.,
(2k + 7) < (k + 3)2 … (1)
We shall now prove that P(k + 1) is true whenever P(k) is true.
Consider
Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.
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: 102n – 1 + 1 is divisible by 11
Prove the following by using the principle of mathematical induction for all n ∈ N: 32n + 2 – 8n– 9 is divisible by 8.
12 + 22 + 32 + ... + n2 =\[\frac{n(n + 1)(2n + 1)}{6}\] .
1 + 3 + 32 + ... + 3n−1 = \[\frac{3^n - 1}{2}\]
\[\frac{1}{2 . 5} + \frac{1}{5 . 8} + \frac{1}{8 . 11} + . . . + \frac{1}{(3n - 1)(3n + 2)} = \frac{n}{6n + 4}\]
\[\frac{1}{3 . 7} + \frac{1}{7 . 11} + \frac{1}{11 . 5} + . . . + \frac{1}{(4n - 1)(4n + 3)} = \frac{n}{3(4n + 3)}\]
1.2 + 2.22 + 3.23 + ... + n.2n = (n − 1) 2n+1+2
1.3 + 2.4 + 3.5 + ... + n. (n + 2) = \[\frac{1}{6}n(n + 1)(2n + 7)\]
1.3 + 3.5 + 5.7 + ... + (2n − 1) (2n + 1) =\[\frac{n(4 n^2 + 6n - 1)}{3}\]
12 + 32 + 52 + ... + (2n − 1)2 = \[\frac{1}{3}n(4 n^2 - 1)\]
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 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:
`[(1, 2),(0, 1)]^"n" = [(1, 2"n"),(0, 1)]` ∀ n ∈ N
Answer the following:
Prove by method of induction 152n–1 + 1 is divisible by 16, for all n ∈ N.
Prove by induction that for all natural number n sinα + sin(α + β) + sin(α + 2β)+ ... + sin(α + (n – 1)β) = `(sin (alpha + (n - 1)/2 beta)sin((nbeta)/2))/(sin(beta/2))`
Let P(n): “2n < (1 × 2 × 3 × ... × n)”. Then the smallest positive integer for which P(n) is true is ______.
Prove the statement by using the Principle of Mathematical Induction:
32n – 1 is divisible by 8, 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:
1 + 2 + 22 + ... + 2n = 2n+1 – 1 for all natural numbers n.
A sequence a1, a2, a3 ... is defined by letting a1 = 3 and ak = 7ak – 1 for all natural numbers k ≥ 2. Show that an = 3.7n–1 for all natural numbers.
For all n ∈ N, 3.52n+1 + 23n+1 is divisible by ______.
If P(n): 2n < n!, n ∈ N, then P(n) is true for all n ≥ ______.
By using principle of mathematical induction for every natural number, (ab)n = ______.
