Advertisements
Advertisements
प्रश्न
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)`
Advertisements
उत्तर
Let P(n) ≡ 12 + 42 + 72 + .... + (3n − 2)2 = `"n"/2 (6"n" − 3"n" − 1)`, for all n ∈ N
Step I:
Put n = 1
L.H.S. = 12 = 1
R.H.S. = `1/2[6(1)^2 - 3(1) - 1]` = 1 = L.H.S.
∴ P(n) is true for n = 1
Step II:
Let us consider that P(n) is true for n = k
∴ 12 + 42 + 72 + .... + (3k − 2)2
= `"k"/2(6"k"^2 - 3"k" - 1)` ...(i)
Step III:
We have to prove that P(n) is true for n = k + 1
i.e., to prove that
12 + 42 + 72 + …. + [3(k + 1) − 2]2
= `(("k" + 1))/3[6("k" + 1)^2 - 3("k" + 1) - 1]`
= `(("k" + 1))/2(6"k"^2 + 12"k" + 6 - 3"k" - 3 - 1)`
= `(("k" + 1))/2 (6"k"^2 + 9"k" + 2)`
L.H.S. = 12 + 42 + 72 + …. + [3(k + 1) − 2]2
= 12 + 42 + 72 + …. + (3k − 2)2 + (3(k + 1) − 2]2
= `"k"/2(6"k"^2 - 3"k" - 1) + (3"k" + 1)^2` ...[From (i)]
= `((6"k"^3 - 3"k"^2 - "k") + 2(9"k"^2 + 6"k" + 1))/2`
= `(6"k"^3 + 15"k"^2 + 11"k" + 2)/2`
= `(("k" + 1)(6"k"^2 + 9"k" + 2))/2`
= 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.
∴ 12 + 42 + 72 + ... + (3n − 2)2 = `"n"/2 (6"n"^2 - 3"n" - 1)` for all n ∈ 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:
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: n (n + 1) (n + 5) is a multiple of 3.
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: 41n – 14n is a multiple of 27.
If P (n) is the statement "n3 + n is divisible by 3", prove that P (3) is true but P (4) is not true.
1 + 2 + 3 + ... + n = \[\frac{n(n + 1)}{2}\] i.e. the sum of the first n natural numbers is \[\frac{n(n + 1)}{2}\] .
1 + 3 + 32 + ... + 3n−1 = \[\frac{3^n - 1}{2}\]
\[\frac{1}{1 . 2} + \frac{1}{2 . 3} + \frac{1}{3 . 4} + . . . + \frac{1}{n(n + 1)} = \frac{n}{n + 1}\]
\[\frac{1}{2 . 5} + \frac{1}{5 . 8} + \frac{1}{8 . 11} + . . . + \frac{1}{(3n - 1)(3n + 2)} = \frac{n}{6n + 4}\]
2 + 5 + 8 + 11 + ... + (3n − 1) = \[\frac{1}{2}n(3n + 1)\]
72n + 23n−3. 3n−1 is divisible by 25 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 n3 - 7n + 3 is divisible by 3 for all n \[\in\] N .
Show by the Principle of Mathematical induction that the sum Sn of then terms of the series \[1^2 + 2 \times 2^2 + 3^2 + 2 \times 4^2 + 5^2 + 2 \times 6^2 + 7^2 + . . .\] is given by \[S_n = \binom{\frac{n \left( n + 1 \right)^2}{2}, \text{ if n is even} }{\frac{n^2 \left( n + 1 \right)}{2}, \text{ if n is odd } }\]
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:
`1/(3.5) + 1/(5.7) + 1/(7.9) + ...` to n terms = `"n"/(3(2"n" + 3))`
Prove by method of induction, for all n ∈ N:
(23n − 1) is divisible by 7
Prove by method of induction, for all n ∈ N:
3n − 2n − 1 is divisible by 4
Answer the following:
Prove, by method of induction, for all n ∈ N
2 + 3.2 + 4.22 + ... + (n + 1)2n–1 = n.2n
Answer the following:
Prove by method of induction 152n–1 + 1 is divisible by 16, 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:
n3 – n is divisible by 6, for each natural number n ≥ 2.
Prove the statement by using the Principle of Mathematical Induction:
n(n2 + 5) is divisible by 6, for each natural number n.
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)`.
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 `1/(n + 1) + 1/(n + 2) + ... + 1/(2n) > 13/24`, for all natural numbers n > 1.
Prove that number of subsets of a set containing n distinct elements is 2n, for all n ∈ N.
Consider the statement: “P(n) : n2 – n + 41 is prime." Then which one of the following is true?
By using principle of mathematical induction for every natural number, (ab)n = ______.
