Advertisements
Advertisements
प्रश्न
12 + 22 + 32 + ... + n2 =\[\frac{n(n + 1)(2n + 1)}{6}\] .
Advertisements
उत्तर
Let P(n) be the given statement.
Now,
\[P(n) = 1^2 + 2^2 + 3^2 + . . . + n^2 = \frac{n(n + 1)(2n + 1)}{6}\]
\[\text{ Step } 1: \]
\[P(1) = 1^2 = \frac{1(1 + 1)(2 + 1)}{6} = \frac{6}{6} = 1\]
\[\text{ Hence, P(1) is true} . \]
\[\text{ Step } 2: \]
\[\text{ Let P(m) be true .} \]
\[\text{ Then,} \]
\[ 1^2 + 2^2 + . . . + m^2 = \frac{m(m + 1)(2m + 1)}{6}\]
\[\text{ We shall now prove that P(m + 1) is true} . \]
\[i . e . , \]
\[ 1^2 + 2^2 + 3^2 + . . . + (m + 1 )^2 = \frac{(m + 1)(m + 2)(2m + 3)}{6}\]
\[ \text{ Now } , \]
\[P(m) = 1^2 + 2^2 + 3^2 + . . . + m^2 = \frac{m(m + 1)(2m + 1)}{6}\]
\[ \Rightarrow 1^2 + 2^2 + 3^2 + . . . + m^2 + (m + 1 )^2 = \frac{m(m + 1)(2m + 1)}{6} + (m + 1 )^2 \left[ \text{ Adding} (m + 1 )^2 \text{ to both sides} \right]\]
\[ \Rightarrow 1^2 + 2^2 + 3^2 + . . . + (m + 1 )^2 = \frac{m(m + 1)(2m + 1) + 6(m + 1 )^2}{6} = \frac{(m + 1)(2 m^2 + m + 6m + 6)}{6} = \frac{(m + 1)(m + 2)(2m + 3)}{6}\]
\[\text{ Hence, P(m + 1) is true } . \]
\[\text{ By the principle of mathematical induction, the given statement is true for all n } \in N .\]
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:
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:
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: 102n – 1 + 1 is divisible by 11
Given an example of a statement P (n) such that it is true for all n ∈ N.
If P (n) is the statement "n2 − n + 41 is prime", prove that P (1), P (2) and P (3) are true. Prove also that P (41) is not true.
1 + 2 + 3 + ... + n = \[\frac{n(n + 1)}{2}\] i.e. the sum of the first n natural numbers is \[\frac{n(n + 1)}{2}\] .
1 + 3 + 32 + ... + 3n−1 = \[\frac{3^n - 1}{2}\]
\[\frac{1}{3 . 7} + \frac{1}{7 . 11} + \frac{1}{11 . 5} + . . . + \frac{1}{(4n - 1)(4n + 3)} = \frac{n}{3(4n + 3)}\]
1.3 + 2.4 + 3.5 + ... + n. (n + 2) = \[\frac{1}{6}n(n + 1)(2n + 7)\]
12 + 32 + 52 + ... + (2n − 1)2 = \[\frac{1}{3}n(4 n^2 - 1)\]
32n+7 is divisible by 8 for all n ∈ N.
52n+2 −24n −25 is divisible by 576 for all n ∈ N.
11n+2 + 122n+1 is divisible by 133 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 \[\frac{a_n - \sqrt{A}}{a_n + \sqrt{A}} = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right) 2^{n - 1}\]
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?
Prove that the number of subsets of a set containing n distinct elements is 2n, for all n \[\in\] N .
\[\text{ A sequence } a_1 , a_2 , a_3 , . . . \text{ is defined by letting } a_1 = 3 \text{ and } a_k = 7 a_{k - 1} \text{ for all natural numbers } k \geq 2 . \text{ Show that } a_n = 3 \cdot 7^{n - 1} \text{ for all } n \in N .\]
\[\text { A sequence } x_1 , x_2 , x_3 , . . . \text{ is defined by letting } x_1 = 2 \text{ and } x_k = \frac{x_{k - 1}}{k} \text{ for all natural numbers } k, k \geq 2 . \text{ Show that } x_n = \frac{2}{n!} \text{ for all } n \in N .\]
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:
12 + 32 + 52 + .... + (2n − 1)2 = `"n"/3 (2"n" − 1)(2"n" + 1)`
Prove by method of induction, for all n ∈ N:
`1/(1.3) + 1/(3.5) + 1/(5.7) + ... + 1/((2"n" - 1)(2"n" + 1)) = "n"/(2"n" + 1)`
Answer the following:
Prove, by method of induction, for all n ∈ N
12 + 42 + 72 + ... + (3n − 2)2 = `"n"/2 (6"n"^2 - 3"n" - 1)`
Answer the following:
Prove, by method of induction, for all n ∈ N
`1/(3.4.5) + 2/(4.5.6) + 3/(5.6.7) + ... + "n"/(("n" + 2)("n" + 3)("n" + 4)) = ("n"("n" + 1))/(6("n" + 3)("n" + 4))`
Prove statement by using the Principle of Mathematical Induction for all n ∈ N, that:
`sum_(t = 1)^(n - 1) t(t + 1) = (n(n - 1)(n + 1))/3`, for all natural numbers n ≥ 2.
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.
State whether the following proof (by mathematical induction) is true or false for the statement.
P(n): 12 + 22 + ... + n2 = `(n(n + 1) (2n + 1))/6`
Proof By the Principle of Mathematical induction, P(n) is true for n = 1,
12 = 1 = `(1(1 + 1)(2*1 + 1))/6`. Again for some k ≥ 1, k2 = `(k(k + 1)(2k + 1))/6`. Now we prove that
(k + 1)2 = `((k + 1)((k + 1) + 1)(2(k + 1) + 1))/6`
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:
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 that, sinθ + sin2θ + sin3θ + ... + sinnθ = `((sin ntheta)/2 sin ((n + 1))/2 theta)/(sin theta/2)`, for all n ∈ N.
For all n ∈ N, 3.52n+1 + 23n+1 is divisible by ______.
