Advertisements
Advertisements
प्रश्न
Answer the following:
Prove, by method of induction, for all n ∈ N
8 + 17 + 26 + … + (9n – 1) = `"n"/2(9"n" + 7)`
Advertisements
उत्तर
Let P(n) ≡ 8 + 17 + 26 + … + (9n – 1) = `"n"/2(9"n" + 7)`, for all n ∈ N
Step I:
Put n = 1
L.H.S. = 8
R.H.S. = `1/2[9(1)+ 7]` = 8 = L.H.S.
∴ P(n) is true for n = 1
Step II:
Let us consider that P(n) is true for n = k
∴ 8 + 17 + 26 + … + (9k – 1) = `"k"/2(9"k" + 7)` ...(i)
Step III:
We have to prove that P(n) is true for n = k + 1
i.e., 8 + 17 + 26 + … + [9(k + 1) – 1]
= `(("k" + 1))/2 [9("k" + 1) + 7]`
= `(("k" + 1))/2 (9"k" + 16)`
L.H.S. = 8 + 17 + 26 + … + [9(k + 1) – 1]
= 8 + 17 + 26 + … + (9k – 1) + [9(k + 1) – 1]
= `"k"/2 (9"k" + 7) + (9"k" + 8)` ...[From (i)]
= `(9"k"^2 + 7"k" + 18"k" + 16)/2`
= `(9"k"^2 + 25"k" + 16)/2`
= `(9"k"^2 + 9"k" + 16"k" + 16)/2`
= `(9"k"("k" + 1) + 16("k" + 1))/2`
= `(("k" + 1))/2(9"k" + 16)`
= R.H.S.
∴ P(n) is true for n = k + 1
Step IV:
From all steps above by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 8 + 17 + 26 + … + (9n – 1) = `"n"/2(9"n" + 7)` for all n ∈ N
APPEARS IN
संबंधित प्रश्न
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:
`(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:
Prove the following by using the principle of mathematical induction for all n ∈ N:
`1/1.4 + 1/4.7 + 1/7.10 + ... + 1/((3n - 2)(3n + 1)) = n/((3n + 1))`
Prove the following by using the principle of mathematical induction for all n ∈ N:
Given an example of a statement P (n) such that it is true for all n ∈ N.
\[\frac{1}{1 . 4} + \frac{1}{4 . 7} + \frac{1}{7 . 10} + . . . + \frac{1}{(3n - 2)(3n + 1)} = \frac{n}{3n + 1}\]
32n+7 is divisible by 8 for all n ∈ N.
52n+2 −24n −25 is divisible by 576 for all n ∈ N.
(ab)n = anbn for all n ∈ N.
11n+2 + 122n+1 is divisible by 133 for all n ∈ N.
Given \[a_1 = \frac{1}{2}\left( a_0 + \frac{A}{a_0} \right), a_2 = \frac{1}{2}\left( a_1 + \frac{A}{a_1} \right) \text{ and } a_{n + 1} = \frac{1}{2}\left( a_n + \frac{A}{a_n} \right)\] for n ≥ 2, where a > 0, A > 0.
Prove that \[\frac{a_n - \sqrt{A}}{a_n + \sqrt{A}} = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right) 2^{n - 1}\]
Prove that 1 + 2 + 22 + ... + 2n = 2n+1 - 1 for all n \[\in\] N .
Let P(n) be the statement : 2n ≥ 3n. If P(r) is true, show that P(r + 1) is true. Do you conclude that P(n) is true for all n ∈ N?
\[\text{ Given } a_1 = \frac{1}{2}\left( a_0 + \frac{A}{a_0} \right), a_2 = \frac{1}{2}\left( a_1 + \frac{A}{a_1} \right) \text{ and } a_{n + 1} = \frac{1}{2}\left( a_n + \frac{A}{a_n} \right) \text{ for } n \geq 2, \text{ where } a > 0, A > 0 . \]
\[\text{ Prove that } \frac{a_n - \sqrt{A}}{a_n + \sqrt{A}} = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right) 2^{n - 1} .\]
Prove that the number of subsets of a set containing n distinct elements is 2n, for all n \[\in\] 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:
(23n − 1) is divisible by 7
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 152n–1 + 1 is divisible by 16, for all n ∈ N.
Answer the following:
Prove by method of induction 52n − 22n is divisible by 3, for all n ∈ N
Let P(n): “2n < (1 × 2 × 3 × ... × n)”. Then the smallest positive integer for which P(n) is true is ______.
Give an example of a statement P(n) which is true for all n ≥ 4 but P(1), P(2) and P(3) are not true. Justify your answer
Prove the statement by using the Principle of Mathematical Induction:
4n – 1 is divisible by 3, for each natural number n.
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:
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:
2n < (n + 2)! for all natural number n.
Prove that, sinθ + sin2θ + sin3θ + ... + sinnθ = `((sin ntheta)/2 sin ((n + 1))/2 theta)/(sin theta/2)`, for all n ∈ N.
Show that `n^5/5 + n^3/3 + (7n)/15` is a natural number for all n ∈ N.
Prove that number of subsets of a set containing n distinct elements is 2n, for all n ∈ N.
