Advertisements
Advertisements
Question
Prove by method of induction, for all n ∈ N:
1.3 + 3.5 + 5.7 + ..... to n terms = `"n"/3(4"n"^2 + 6"n" - 1)`
Advertisements
Solution
Let P(n) ≡ 1.3 + 3.5 + 5.7 + ..... to n terms = `"n"/3(4"n"^2 + 6"n" - 1)`, for all n ∈ N
But the first factor in each term
i.e., 1, 3, 5 … are in A.P. with a = 1 and d = 2.
∴ nth term = a + (n –1)d = 1 +(n – 1)2 = (2n – 1)
Also second factor in each term
i.e., 3, 5, 7, … are in A.P. with a = 3 and d = 2.
∴ nth term = a + (n – 1)d = 3 + (n – 1)2 = (2n+1)
∴ nth term, tn = (2n – 1) (2n + 1)
∴ P(n) ≡ 1.3 + 3.5 + 5.7 + .... + (2n – 1) (2n + 1) = `"n"/3(4"n"^2 + 6"n" - 1)`
Step I:
Put n = 1
L.H.S. = 1.3 = 3
R.H.S. = `1/3[4(1)^2 + 6(1) - 1]` = 3 = L.H.S.
∴ P(n) is true for n = 1.
Step II:
Let us consider that P(n) is true for n = k
∴ 1.3 + 3.5 + 5.7 + ..... + (2k – 1)(2k + 1)
= `"k"/3(4"k"^2 + 6"k" - 1)` ...(i)
Step III:
We have to prove that P(n) is true for n = k + 1
i.e., to prove that
1.3 + 3.5 + 5.7 + …. + [2(k + 1) – 1][2(k + 1) + 1]
= `(("k" + 1))/3[4("k" + 1)^2 + 6("k" + 1) - 1]`
= `(("k" + 1))/3(4"k"^2 + 8"k" + 4 + 6"k" + 6 - 1)`
= `(("k" + 1))/3(4"k"^2 + 14"k" + 9)`
L.H.S. = 1.3 + 3.5 + 5.7 + ... + [2(k + 1) – 1][2(k + 1) + 1]
= 1.3 + 3.5 + 5.7 + ... + (2k – 1)(2k + 1) + (2k + 1)(2k + 3)
= `"k"/3(4"k"^2 + 6"k" - 1) + (2"k" + 1) (2"k" + 3)` ...[From (i)]
= `1/3[4"k"^3 + 6"k"^2 - "k" + 3(2"k" + 1)(2"k" + 3)]`
= `1/3(4"k"^3 + 6"k"^2 - "k" + 12"k"^2 + 24"k" + 9)`
= `1/3(4"k"^3 + 18"k"^2 + 23"k" + 9)`
= `1/3("k" + 1)(4"k"^2 + 14"k" + 9)`
= 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.
∴ 1.3 + 3.5 + 5.7 + ..... to n terms = `"n"/3(4"n"^2 + 6"n" - 1)` for all n ∈ N.
APPEARS IN
RELATED QUESTIONS
Prove the following by using the principle of mathematical induction for all n ∈ N:
`1 + 3 + 3^2 + ... + 3^(n – 1) =((3^n -1))/2`
Prove the following by using the principle of mathematical induction for all n ∈ N:
1.2 + 2.3 + 3.4+ ... + n(n+1) = `[(n(n+1)(n+2))/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:
Prove the following by using the principle of mathematical induction for all n ∈ N: `1+2+ 3+...+n<1/8(2n +1)^2`
Prove the following by using the principle of mathematical induction for all n ∈ N: 32n + 2 – 8n– 9 is divisible by 8.
Given an example of a statement P (n) such that it is true for all n ∈ N.
12 + 22 + 32 + ... + n2 =\[\frac{n(n + 1)(2n + 1)}{6}\] .
\[\frac{1}{2 . 5} + \frac{1}{5 . 8} + \frac{1}{8 . 11} + . . . + \frac{1}{(3n - 1)(3n + 2)} = \frac{n}{6n + 4}\]
2 + 5 + 8 + 11 + ... + (3n − 1) = \[\frac{1}{2}n(3n + 1)\]
1.3 + 3.5 + 5.7 + ... + (2n − 1) (2n + 1) =\[\frac{n(4 n^2 + 6n - 1)}{3}\]
12 + 32 + 52 + ... + (2n − 1)2 = \[\frac{1}{3}n(4 n^2 - 1)\]
52n+2 −24n −25 is divisible by 576 for all n ∈ N.
11n+2 + 122n+1 is divisible by 133 for all n ∈ N.
\[\text{ Given } a_1 = \frac{1}{2}\left( a_0 + \frac{A}{a_0} \right), a_2 = \frac{1}{2}\left( a_1 + \frac{A}{a_1} \right) \text{ and } a_{n + 1} = \frac{1}{2}\left( a_n + \frac{A}{a_n} \right) \text{ for } n \geq 2, \text{ where } a > 0, A > 0 . \]
\[\text{ Prove that } \frac{a_n - \sqrt{A}}{a_n + \sqrt{A}} = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right) 2^{n - 1} .\]
\[\text{ Let } P\left( n \right) \text{ be the statement } : 2^n \geq 3n . \text{ If } P\left( r \right) \text{ is true, then show that } P\left( r + 1 \right) \text{ is true . Do you conclude that } P\left( n \right)\text{ is true for all n } \in N?\]
\[\text{ A sequence } a_1 , a_2 , a_3 , . . . \text{ is defined by letting } a_1 = 3 \text{ and } a_k = 7 a_{k - 1} \text{ for all natural numbers } k \geq 2 . \text{ Show that } a_n = 3 \cdot 7^{n - 1} \text{ for all } n \in N .\]
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:
12 + 32 + 52 + .... + (2n − 1)2 = `"n"/3 (2"n" − 1)(2"n" + 1)`
Prove by method of induction, for all n ∈ N:
5 + 52 + 53 + .... + 5n = `5/4(5^"n" - 1)`
Prove by method of induction, for all n ∈ N:
`[(1, 2),(0, 1)]^"n" = [(1, 2"n"),(0, 1)]` ∀ n ∈ N
Prove statement by using the Principle of Mathematical Induction for all n ∈ N, that:
`(1 - 1/2^2).(1 - 1/3^2)...(1 - 1/n^2) = (n + 1)/(2n)`, for all natural numbers, n ≥ 2.
Prove statement by using the Principle of Mathematical Induction for all n ∈ N, that:
22n – 1 is divisible by 3.
Define the sequence a1, a2, a3 ... as follows:
a1 = 2, an = 5 an–1, for all natural numbers n ≥ 2.
Use the Principle of Mathematical Induction to show that the terms of the sequence satisfy the formula an = 2.5n–1 for all natural numbers.
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 – 7n + 3 is divisible by 3, for all natural numbers n.
Prove the statement by using the Principle of Mathematical Induction:
1 + 5 + 9 + ... + (4n – 3) = n(2n – 1) for all natural numbers n.
A sequence b0, b1, b2 ... is defined by letting b0 = 5 and bk = 4 + bk – 1 for all natural numbers k. Show that bn = 5 + 4n for all natural number n using mathematical induction.
If 10n + 3.4n+2 + k is divisible by 9 for all n ∈ N, then the least positive integral value of k is ______.
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 ≥ ______.
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.
