Advertisements
Advertisements
Question
\[\frac{1}{1 . 4} + \frac{1}{4 . 7} + \frac{1}{7 . 10} + . . . + \frac{1}{(3n - 2)(3n + 1)} = \frac{n}{3n + 1}\]
Advertisements
Solution
Let P(n) be the given statement.
Now,
\[P(n) = \frac{1}{1 . 4} + \frac{1}{4 . 7} + \frac{1}{7 . 10} + . . . + \frac{1}{(3n - 2)(3n + 1)} = \frac{n}{3n + 1}\]
\[\text{ Step } 1: \]
\[P(1) = \frac{1}{1 . 4} = \frac{1}{4} = \frac{1}{3 \times 1 + 1}\]
\[\text{ Hence, P(1) is true } . \]
\[Step 2: \]
\[\text{ Let P(m) be true} . \]
\[i . e . , \]
\[\frac{1}{1 . 4} + \frac{1}{4 . 7} + . . . + \frac{1}{(3m - 2)(3m + 1)} = \frac{m}{3m + 1}\]
\[\text{ To prove: P(m + 1) is true } . \]
\[i . e . , \]
\[\frac{1}{1 . 4} + \frac{1}{4 . 7} + . . . + \frac{1}{(3m + 1)(3m + 4)} = \frac{m + 1}{3m + 4}\]
\[\text{ Now, } \]
\[P(m) = \frac{1}{1 . 4} + \frac{1}{4 . 7} + . . . + \frac{1}{(3m - 2)(3m + 1)} = \frac{m}{3m + 1}\]
\[ \Rightarrow \frac{1}{1 . 4} + \frac{1}{4 . 7} + . . . + \frac{1}{(3m - 2)(3m + 1)} + \frac{1}{(3m + 1)(3m + 4)} = \frac{m}{3m + 1} + \frac{1}{(3m + 1)(3m + 4)} \left[ \text{ Adding } \frac{1}{(3m + 1)(3m + 4)} \text{ to both sides} \right]\]
\[ \Rightarrow \frac{1}{1 . 4} + \frac{1}{4 . 7} + . . . + \frac{1}{(3m + 1)(3m + 4)} = \frac{3 m^2 + 4m + 1}{(3m + 1)(3m + 4)} = \frac{(3m + 1)(m + 1)}{(3m + 1)(3m + 4)} = \frac{m + 1}{3m + 4}\]
\[\text{ Thus, P(m + 1) is true .} \]
\[\text{ By the principle of mathematical induction, P(n) is true for all n} \in N .\]
APPEARS IN
RELATED QUESTIONS
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:
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
Given an example of a statement P (n) such that it is true for all n ∈ N.
12 + 22 + 32 + ... + n2 =\[\frac{n(n + 1)(2n + 1)}{6}\] .
2 + 5 + 8 + 11 + ... + (3n − 1) = \[\frac{1}{2}n(3n + 1)\]
1.3 + 2.4 + 3.5 + ... + n. (n + 2) = \[\frac{1}{6}n(n + 1)(2n + 7)\]
12 + 32 + 52 + ... + (2n − 1)2 = \[\frac{1}{3}n(4 n^2 - 1)\]
52n −1 is divisible by 24 for all n ∈ N.
72n + 23n−3. 3n−1 is divisible by 25 for all n ∈ N.
Prove that 1 + 2 + 22 + ... + 2n = 2n+1 - 1 for all n \[\in\] N .
7 + 77 + 777 + ... + 777 \[{. . . . . . . . . . .}_{n - \text{ digits } } 7 = \frac{7}{81}( {10}^{n + 1} - 9n - 10)\]
\[\text{ Given } a_1 = \frac{1}{2}\left( a_0 + \frac{A}{a_0} \right), a_2 = \frac{1}{2}\left( a_1 + \frac{A}{a_1} \right) \text{ and } a_{n + 1} = \frac{1}{2}\left( a_n + \frac{A}{a_n} \right) \text{ for } n \geq 2, \text{ where } a > 0, A > 0 . \]
\[\text{ Prove that } \frac{a_n - \sqrt{A}}{a_n + \sqrt{A}} = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right) 2^{n - 1} .\]
Prove 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:
(23n − 1) is divisible by 7
Prove by method of induction, for all n ∈ N:
3n − 2n − 1 is divisible by 4
Prove by method of induction, for all n ∈ N:
Given that tn+1 = 5tn + 4, t1 = 4, prove that tn = 5n − 1
Answer the following:
Given that tn+1 = 5tn − 8, t1 = 3, prove by method of induction that tn = 5n−1 + 2
Answer the following:
Prove by method of induction 152n–1 + 1 is divisible by 16, for all n ∈ N.
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:
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:
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:
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 the statement by using the Principle of Mathematical Induction:
1 + 2 + 22 + ... + 2n = 2n+1 – 1 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 a1, a2, a3 ... is defined by letting a1 = 3 and ak = 7ak – 1 for all natural numbers k ≥ 2. Show that an = 3.7n–1 for all natural numbers.
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.
Prove that, cosθ cos2θ cos22θ ... cos2n–1θ = `(sin 2^n theta)/(2^n sin theta)`, for all n ∈ N.
Prove that `1/(n + 1) + 1/(n + 2) + ... + 1/(2n) > 13/24`, for all natural numbers n > 1.
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.
By using principle of mathematical induction for every natural number, (ab)n = ______.
