Advertisements
Advertisements
प्रश्न
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 \[S_n = \binom{\frac{n \left( n + 1 \right)^2}{2}, \text{ if n is even} }{\frac{n^2 \left( n + 1 \right)}{2}, \text{ if n is odd } }\]
Advertisements
उत्तर
\[ \text{ Let } P\left( n \right): S_n = 1^2 + 2 \times 2^2 + 3^2 + 2 \times 4^2 + 5^2 + . . . = \binom{\frac{n \left( n + 1 \right)^2}{2}, \text{ when n is even } }{\frac{n^2 \left( n + 1 \right)}{2}, \text{ when n is odd} }\]
\[\text{ Step I: For } n = 1 i . e . P\left( 1 \right): \]
\[LHS = S_1 = 1^2 = 1\]
\[RHS = S_1 = \frac{1^2 \left( 1 + 1 \right)}{2} = 1\]
\[\text{ As, LHS = RHS } \]
\[\text{ So, it is true for n } = 1 . \]
\[ \text{ Step II: For n } = k, \]
\[\text{ Let } P\left( k \right): S_k = 1^2 + 2 \times 2^2 + 3^2 + 2 \times 4^2 + 5^2 + . . . = \binom{\frac{k \left( k + 1 \right)^2}{2}, \text{ when k is even} }{\frac{k^2 \left( k + 1 \right)}{2}, \text{ when k is odd }}, \text{ be true for some natural numbers } . \]
\[\text{ Step III: For } n = k + 1, \]
\[\text{ Case 1: When k is odd, then } \left( k + 1 \right) \text{ is even .} \]
\[P\left( k + 1 \right): \]
\[LHS = S_{k + 1} = 1^2 + 2 \times 2^2 + 3^2 + 2 \times 4^2 + 5^2 + . . . + k^2 + 2 \times \left( k + 1 \right)^2 \]
\[ = \frac{k^2 \left( k + 1 \right)}{2} + 2 \times \left( k + 1 \right)^2 \left( \text{ Using step } II \right)\]
\[ = \frac{k^2 \left( k + 1 \right) + 4 \left( k + 1 \right)^2}{2}\]
\[ = \frac{\left( k + 1 \right)\left( k^2 + 4k + 4 \right)}{2}\]
\[ = \frac{\left( k + 1 \right) \left( k + 2 \right)^2}{2}\]
\[RHS = \frac{\left( k + 1 \right) \left( k + 1 + 1 \right)^2}{2}\]
\[ = \frac{\left( k + 1 \right) \left( k + 2 \right)^2}{2}\]
\[\text{ As, LHS = RHS } \]
\[\text{ So, it is true for n = k + 1 when k is odd .} \]
\[\text{ Case 2: When k is even, then } \left( k + 1 \right) \text{ is odd .} \]
\[P\left( k + 1 \right): \]
\[RHS = S_{k + 1} = 1^2 + 2 \times 2^2 + 3^2 + 2 \times 4^2 + 5^2 + . . . + 2 \times k^2 + \left( k + 1 \right)^2 \]
\[ = \frac{k \left( k + 1 \right)^2}{2} + \left( k + 1 \right)^2 \left( \text{ Using step } II \right)\]
\[ = \frac{k \left( k + 1 \right)^2 + 2 \left( k + 1 \right)^2}{2}\]
\[ = \frac{\left( k + 1 \right)^2 \left( k + 2 \right)}{2}\]
\[ = \frac{\left( k + 1 \right)^2 \left( k + 2 \right)}{2}\]
\[RHS = \frac{\left( k + 1 \right)^2 \left( k + 1 + 1 \right)}{2}\]
\[ = \frac{\left( k + 1 \right)^2 \left( k + 2 \right)}{2}\]
\[As, LHS = RHS\]
\[\text{ So, it is true for n = k + 1 when k is even } . \]
\[\text{ Hence, } P\left( n \right) \text{ is true for all natural numbers .} \]
APPEARS IN
संबंधित प्रश्न
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+ 1/1)(1+ 1/2)(1+ 1/3)...(1+ 1/n) = (n + 1)`
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: 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 "n2 + n is even", and if P (r) is true, then P (r + 1) is true.
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.
12 + 22 + 32 + ... + n2 =\[\frac{n(n + 1)(2n + 1)}{6}\] .
\[\frac{1}{1 . 2} + \frac{1}{2 . 3} + \frac{1}{3 . 4} + . . . + \frac{1}{n(n + 1)} = \frac{n}{n + 1}\]
\[\frac{1}{2 . 5} + \frac{1}{5 . 8} + \frac{1}{8 . 11} + . . . + \frac{1}{(3n - 1)(3n + 2)} = \frac{n}{6n + 4}\]
\[\frac{1}{1 . 4} + \frac{1}{4 . 7} + \frac{1}{7 . 10} + . . . + \frac{1}{(3n - 2)(3n + 1)} = \frac{n}{3n + 1}\]
1.3 + 2.4 + 3.5 + ... + n. (n + 2) = \[\frac{1}{6}n(n + 1)(2n + 7)\]
1.3 + 3.5 + 5.7 + ... + (2n − 1) (2n + 1) =\[\frac{n(4 n^2 + 6n - 1)}{3}\]
a + (a + d) + (a + 2d) + ... (a + (n − 1) d) = \[\frac{n}{2}\left[ 2a + (n - 1)d \right]\]
52n+2 −24n −25 is divisible by 576 for all n ∈ N.
32n+2 −8n − 9 is divisible by 8 for all n ∈ N.
2.7n + 3.5n − 5 is divisible by 24 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 the number of subsets of a set containing n distinct elements is 2n, for all n \[\in\] N .
\[\text{ A sequence } x_0 , x_1 , x_2 , x_3 , . . . \text{ is defined by letting } x_0 = 5 and x_k = 4 + x_{k - 1}\text{ for all natural number k . } \]
\[\text{ Show that } x_n = 5 + 4n \text{ for all n } \in N \text{ using mathematical induction .} \]
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)`
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
8 + 17 + 26 + … + (9n – 1) = `"n"/2(9"n" + 7)`
Answer the following:
Prove by method of induction 152n–1 + 1 is divisible by 16, for all n ∈ N.
Answer the following:
Prove by method of induction 52n − 22n is divisible by 3, for all n ∈ N
Prove statement by using the Principle of Mathematical Induction for all n ∈ N, that:
1 + 3 + 5 + ... + (2n – 1) = n2
Show by the Principle of Mathematical Induction that the sum Sn of the n term of the series 12 + 2 × 22 + 32 + 2 × 42 + 52 + 2 × 62 ... is given by
Sn = `{{:((n(n + 1)^2)/2",", "if n is even"),((n^2(n + 1))/2",", "if n is odd"):}`
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:
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:
n3 – 7n + 3 is divisible by 3, for all natural numbers 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:
2n < (n + 2)! for all natural number 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.
If 10n + 3.4n+2 + k is divisible by 9 for all n ∈ N, then the least positive integral value of k is ______.
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.
