Advertisements
Advertisements
Question
\[\frac{1}{2 . 5} + \frac{1}{5 . 8} + \frac{1}{8 . 11} + . . . + \frac{1}{(3n - 1)(3n + 2)} = \frac{n}{6n + 4}\]
Advertisements
Solution
Let P(n) be the given statement.
Now,
\[P(n) = \frac{1}{2 . 5} + \frac{1}{5 . 8} + \frac{1}{8 . 11} + . . . + \frac{1}{(3n - 1)(3n + 2)} = \frac{n}{6n + 4}\]
\[\text{ Step } 1: \]
\[P(1) = \frac{1}{2 . 5} = \frac{1}{10} = \frac{1}{6 + 4}\]
\[\text{ Hence, P(1) is true } . \]
\[\text{ Step 2:} \]
\[\text{ Let P(m) be true .} \]
\[\text{ Then,} \]
\[\frac{1}{2 . 5} + \frac{1}{5 . 8} + \frac{1}{8 . 11} + . . . + \frac{1}{(3m - 1)(3m + 2)} = \frac{m}{6m + 4}\]
\[\text{ To prove: P(m + 1) is true } . \]
\[i . e . , \]
\[\frac{1}{2 . 5} + \frac{1}{5 . 8} + . . . + \frac{1}{(3m + 2)(3m + 5)} = \frac{m + 1}{6m + 10}\]
\[\text{ Thus, we have }: \]
\[ \frac{1}{2 . 5} + \frac{1}{5 . 8} + \frac{1}{8 . 11} + . . . + \frac{1}{(3m - 1)(3m + 2)} = \frac{m}{6m + 4}\]
\[ \Rightarrow \frac{1}{2 . 5} + \frac{1}{5 . 8} + . . . + \frac{1}{(3m - 1)(3m + 2)} + \frac{1}{(3m + 2)(3m + 5)} = \frac{m}{6m + 4} + \frac{1}{(3m + 2)(3m + 5)} \left[ \text{ Adding } \frac{1}{(3m + 2)(3m + 5)} \text{ to both sides } \right]\]
\[ \Rightarrow \frac{1}{2 . 5} + \frac{1}{5 . 8} + . . . + \frac{1}{(3m + 2)(3m + 5)} = \frac{3 m^2 + 5m + 2}{2(3m + 2)(3m + 5)} = \frac{(3m + 2)(m + 1)}{2(3m + 2)(3m + 5)} = \frac{m + 1}{6m + 10}\]
\[\text{ Thus, P(m + 1) is true } . \]
\[\text{ By the principle of mathematical induction, P(n) is true for all n } \in N .\]
APPEARS IN
RELATED QUESTIONS
Prove the following by using the principle of mathematical induction for all n ∈ N:
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+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: x2n – y2n is divisible by x + y.
Prove the following by using the principle of mathematical induction for all n ∈ N (2n +7) < (n + 3)2
If P (n) is the statement "n(n + 1) is even", then what is P(3)?
If P (n) is the statement "n2 + n is even", and if P (r) is true, then P (r + 1) is true.
Given an example of a statement P (n) such that it is true for all n ∈ N.
1.2 + 2.3 + 3.4 + ... + n (n + 1) = \[\frac{n(n + 1)(n + 2)}{3}\]
a + ar + ar2 + ... + arn−1 = \[a\left( \frac{r^n - 1}{r - 1} \right), r \neq 1\]
52n+2 −24n −25 is divisible by 576 for all n ∈ N.
(ab)n = anbn for all n ∈ 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?
\[\frac{(2n)!}{2^{2n} (n! )^2} \leq \frac{1}{\sqrt{3n + 1}}\] for all n ∈ N .
\[\text{ Given } a_1 = \frac{1}{2}\left( a_0 + \frac{A}{a_0} \right), a_2 = \frac{1}{2}\left( a_1 + \frac{A}{a_1} \right) \text{ and } a_{n + 1} = \frac{1}{2}\left( a_n + \frac{A}{a_n} \right) \text{ for } n \geq 2, \text{ where } a > 0, A > 0 . \]
\[\text{ Prove that } \frac{a_n - \sqrt{A}}{a_n + \sqrt{A}} = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right) 2^{n - 1} .\]
\[\text{ Let } P\left( n \right) \text{ be the statement } : 2^n \geq 3n . \text{ If } P\left( r \right) \text{ is true, then show that } P\left( r + 1 \right) \text{ is true . Do you conclude that } P\left( n \right)\text{ is true for all n } \in N?\]
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:
13 + 33 + 53 + .... to n terms = n2(2n2 − 1)
Prove by method of induction, for all n ∈ N:
1.2 + 2.3 + 3.4 + ..... + n(n + 1) = `"n"/3 ("n" + 1)("n" + 2)`
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)`
Answer the following:
Prove by method of induction
`[(3, -4),(1, -1)]^"n" = [(2"n" + 1, -4"n"),("n", -2"n" + 1)], ∀ "n" ∈ "N"`
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. Justify your answer.
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, 7n – 2n is divisible by 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 + 5 + 9 + ... + (4n – 3) = n(2n – 1) for all natural numbers n.
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.
Prove that number of subsets of a set containing n distinct elements is 2n, for all n ∈ N.
If P(n): 2n < n!, n ∈ N, then P(n) is true for all n ≥ ______.
