Advertisements
Advertisements
Question
Prove by method of induction, for all n ∈ N:
12 + 32 + 52 + .... + (2n − 1)2 = `"n"/3 (2"n" − 1)(2"n" + 1)`
Solution
Let P(n) ≡ 12 + 32 + 52 + .... + (2n − 1)2 = `"n"/3 (2"n" − 1)(2"n" + 1)`, for all n ∈ N
Step I:
Put n = 1
L.H.S. = 12 = 1
R.H.S. = `1/3[2(1) - 1][(2(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 + 32 + 52 + .... + (2k − 1)2
= `"k"/3(2"k" - 1) (2"k" + 1)` ...(i)
Step III:
We have to prove that P(n) is true for n = k + 1
i.e., to prove that
12 + 32 + 52 + …. + [2(k + 1) − 1]2
= `(("k" + 1))/3[2("k" + 1) - 1][2("k" + 1) + 1]`
= `(("k" + 1)(2"k" + 1)(2"k" + 3))/3`
L.H.S. = 12 + 32 + 52 + …. + [2(k + 1) − 1]2
= 12 + 32 + 52 + …. + (2k − 1)2 + (2k + 1)2
= `"k"/3(2"k" - 1)(2"k" + 1) + (2"k" + 1)^2` ...[From (i)]
= `(2"k" + 1)[("k"(2"k" - 1))/3 + (2"k" + 1)]`
= `(2"k" + 1)[(2"k"^2 - "k" + 6"k" + 3)/3]`
= `((2"k" + 1))/3(2"k"^2 + 2"k" + 3"k" + 3)`
= `((2"k" + 1))/3[2"k"("k" + 1) + 3("k" + 1)]`
= `((2"k" + 1)("k" + 1)(2"k" + 3))/3`
= 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 + 32 + 52 + .... + (2n − 1)2 = `"n"/3 (2"n" − 1)(2"n" + 1)`, for all n ∈ N.
APPEARS IN
RELATED QUESTIONS
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+3/1)(1+ 5/4)(1+7/9)...`(1 + ((2n + 1))/n^2) = (n + 1)^2`
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: x2n – y2n is divisible by x + y.
Prove the following by using the principle of mathematical induction for all n ∈ N: 32n + 2 – 8n– 9 is divisible by 8.
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.
Given an example of a statement P (n) such that it is true for all n ∈ N.
1 + 2 + 3 + ... + n = \[\frac{n(n + 1)}{2}\] i.e. the sum of the first n natural numbers is \[\frac{n(n + 1)}{2}\] .
\[\frac{1}{1 . 4} + \frac{1}{4 . 7} + \frac{1}{7 . 10} + . . . + \frac{1}{(3n - 2)(3n + 1)} = \frac{n}{3n + 1}\]
\[\frac{1}{3 . 5} + \frac{1}{5 . 7} + \frac{1}{7 . 9} + . . . + \frac{1}{(2n + 1)(2n + 3)} = \frac{n}{3(2n + 3)}\]
\[\frac{1}{3 . 7} + \frac{1}{7 . 11} + \frac{1}{11 . 5} + . . . + \frac{1}{(4n - 1)(4n + 3)} = \frac{n}{3(4n + 3)}\]
32n+2 −8n − 9 is divisible by 8 for all n ∈ N.
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 1 + 2 + 22 + ... + 2n = 2n+1 - 1 for all n \[\in\] N .
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:
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:
12 + 22 + 32 + .... + n2 = `("n"("n" + 1)(2"n" + 1))/6`
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
Answer the following:
Prove, by method of induction, for all n ∈ N
8 + 17 + 26 + … + (9n – 1) = `"n"/2(9"n" + 7)`
Answer the following:
Given that tn+1 = 5tn − 8, t1 = 3, prove by method of induction that tn = 5n−1 + 2
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:
For any natural number n, xn – yn is divisible by x – y, where x and y are any integers with x ≠ y.
Prove the statement by using the Principle of Mathematical Induction:
n(n2 + 5) is divisible by 6, for each natural number 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 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 number of subsets of a set containing n distinct elements is 2n, for all n ∈ N.
For all n ∈ N, 3.52n+1 + 23n+1 is divisible by ______.
State whether the following statement is true or false. Justify.
Let P(n) be a statement and let P(k) ⇒ P(k + 1), for some natural number k, then P(n) is true for all n ∈ N.
