Advertisements
Advertisements
प्रश्न
Prove the statement by using the Principle of Mathematical Induction:
For any natural number n, 7n – 2n is divisible by 5.
Advertisements
उत्तर
P(n) = 7n – 2n is divisible by 5.
So, substituting different values for n, we get,
P(0) = 70 – 20 = 0 Which is divisible by 5.
P(1) = 71 – 21 = 5 Which is divisible by 5.
P(2) = 72 – 22 = 45 Which is divisible by 5.
P(3) = 73 – 23 = 335 Which is divisible by 5.
Let P(k) = 7k – 2k be divisible by 5.
So, we get,
⇒ 7k – 2k = 5x
Now, we also get that,
⇒ P(k + 1)= 7k+1 – 2k+1
= (5 + 2)7k – 2(2k)
= 5(7k) + 2(7k – 2k)
= 5(7k) + 2(5x) Which is divisible by 5.
⇒ P(k + 1) is true when P(k) is true.
Therefore, by Mathematical Induction, P(n) = 7n – 2n is divisible by 5 is true for each natural number n.
APPEARS IN
संबंधित प्रश्न
Prove the following by using the principle of mathematical induction for all n ∈ N:
`1^3 + 2^3 + 3^3 + ... + n^3 = ((n(n+1))/2)^2`
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:
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: 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+ 1/1)(1+ 1/2)(1+ 1/3)...(1+ 1/n) = (n + 1)`
Prove the following by using the principle of mathematical induction for all n ∈ N: 41n – 14n is a multiple of 27.
If P (n) is the statement "2n ≥ 3n" and if P (r) is true, prove that P (r + 1) is true.
If P (n) is the statement "n2 − n + 41 is prime", prove that P (1), P (2) and P (3) are true. Prove also that P (41) is not true.
12 + 22 + 32 + ... + n2 =\[\frac{n(n + 1)(2n + 1)}{6}\] .
1.2 + 2.22 + 3.23 + ... + n.2n = (n − 1) 2n+1+2
1.3 + 2.4 + 3.5 + ... + n. (n + 2) = \[\frac{1}{6}n(n + 1)(2n + 7)\]
1.2 + 2.3 + 3.4 + ... + n (n + 1) = \[\frac{n(n + 1)(n + 2)}{3}\]
a + ar + ar2 + ... + arn−1 = \[a\left( \frac{r^n - 1}{r - 1} \right), r \neq 1\]
a + (a + d) + (a + 2d) + ... (a + (n − 1) d) = \[\frac{n}{2}\left[ 2a + (n - 1)d \right]\]
52n −1 is divisible by 24 for all n ∈ N.
2.7n + 3.5n − 5 is divisible by 24 for all n ∈ N.
Prove that 1 + 2 + 22 + ... + 2n = 2n+1 - 1 for all n \[\in\] N .
\[\frac{(2n)!}{2^{2n} (n! )^2} \leq \frac{1}{\sqrt{3n + 1}}\] for all n ∈ N .
Prove by method of induction, for all n ∈ N:
2 + 4 + 6 + ..... + 2n = n (n+1)
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:
(24n−1) is divisible by 15
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 52n − 22n is divisible by 3, for all n ∈ N
Prove statement by using the Principle of Mathematical Induction for all n ∈ N, that:
1 + 3 + 5 + ... + (2n – 1) = n2
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))`
Prove by the Principle of Mathematical Induction that 1 × 1! + 2 × 2! + 3 × 3! + ... + n × n! = (n + 1)! – 1 for all natural numbers n.
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:
For any natural number n, xn – yn is divisible by x – y, where x and y are any integers with x ≠ y.
A sequence d1, d2, d3 ... is defined by letting d1 = 2 and dk = `(d_(k - 1))/"k"` for all natural numbers, k ≥ 2. Show that dn = `2/(n!)` for all n ∈ N.
Prove that `1/(n + 1) + 1/(n + 2) + ... + 1/(2n) > 13/24`, for all natural numbers n > 1.
