Advertisements
Advertisements
प्रश्न
2 + 5 + 8 + 11 + ... + (3n − 1) = \[\frac{1}{2}n(3n + 1)\]
उत्तर
Let P(n) be the given statement.
Now,
\[P(n) = 2 + 5 + 8 + . . . + (3n - 1) = \frac{1}{2}n(3n + 1)\]
\[\text{ Step} 1: \]
\[P(1) = 2 = \frac{1}{2} \times 1(3 + 1) \]
\[\text{ Hence, P(1) is true .} \]
\[\text{ Step} 2: \]
\[\text{ Let P(m) be true } . \]
\[\text{ Then, } \]
\[2 + 5 + 8 + . . . + (3m - 1) = \frac{1}{2}m(3m + 1)\]
\[\text{ To prove: P(m + 1) is true } . \]
\[\text{ That is, } \]
\[2 + 5 + 8 + . . . + (3m + 2) = \frac{1}{2}(m + 1)(3m + 4)\]
\[\text{ P(m) is equal to:} \]
\[2 + 5 + 8 + . . . + (3m - 1) = \frac{1}{2}m(3m + 1)\]
\[\text{ Thus, we have:} \]
\[2 + 5 + 8 + . . . + (3m - 1) + (3m + 2) = \frac{1}{2}m(3m + 1) + (3m + 2) \left[ \text{ Adding } (3m + 2) \text{ to both sides } ] \right]\]
\[ \Rightarrow 2 + 5 + 8 + . . . + (3m + 2) = \frac{1}{2}(3 m^2 + m + 6m + 4) = \frac{1}{2}(3 m^2 + 7m + 4)\]
\[ \Rightarrow 2 + 5 + 8 + . . . + (3m + 2) = \frac{1}{2}(3m + 4)(m + 1)\]
\[\text{ Thus, P(m + 1) is true } . \]
\[\text{ By the principle of mathematical induction, P(n) is true for all n} \in N .\]
APPEARS IN
संबंधित प्रश्न
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:
Prove the following by using the principle of mathematical induction for all n ∈ N: `1/2 + 1/4 + 1/8 + ... + 1/2^n = 1 - 1/2^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: n (n + 1) (n + 5) is a multiple of 3.
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 "n(n + 1) is even", then what is P(3)?
If P (n) is the statement "n3 + n is divisible by 3", prove that P (3) is true but P (4) is not true.
If P (n) is the statement "n2 + n is even", and if P (r) is true, then P (r + 1) is true.
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.
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.
\[\frac{1}{1 . 2} + \frac{1}{2 . 3} + \frac{1}{3 . 4} + . . . + \frac{1}{n(n + 1)} = \frac{n}{n + 1}\]
\[\frac{1}{3 . 5} + \frac{1}{5 . 7} + \frac{1}{7 . 9} + . . . + \frac{1}{(2n + 1)(2n + 3)} = \frac{n}{3(2n + 3)}\]
\[\frac{1}{3 . 7} + \frac{1}{7 . 11} + \frac{1}{11 . 5} + . . . + \frac{1}{(4n - 1)(4n + 3)} = \frac{n}{3(4n + 3)}\]
(ab)n = anbn for all n ∈ N.
72n + 23n−3. 3n−1 is divisible by 25 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 \[\frac{a_n - \sqrt{A}}{a_n + \sqrt{A}} = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right) 2^{n - 1}\]
\[\text{ Let } P\left( n \right) \text{ be the statement } : 2^n \geq 3n . \text{ If } P\left( r \right) \text{ is true, then show that } P\left( r + 1 \right) \text{ is true . Do you conclude that } P\left( n \right)\text{ is true for all n } \in N?\]
\[\text{ A sequence } a_1 , a_2 , a_3 , . . . \text{ is defined by letting } a_1 = 3 \text{ and } a_k = 7 a_{k - 1} \text{ for all natural numbers } k \geq 2 . \text{ Show that } a_n = 3 \cdot 7^{n - 1} \text{ for all } n \in N .\]
\[\text{ A sequence } x_0 , x_1 , x_2 , x_3 , . . . \text{ is defined by letting } x_0 = 5 and x_k = 4 + x_{k - 1}\text{ for all natural number k . } \]
\[\text{ Show that } x_n = 5 + 4n \text{ for all n } \in N \text{ using mathematical induction .} \]
\[\text{ The distributive law from algebra states that for all real numbers} c, a_1 \text{ and } a_2 , \text{ we have } c\left( a_1 + a_2 \right) = c a_1 + c a_2 . \]
\[\text{ Use this law and mathematical induction to prove that, for all natural numbers, } n \geq 2, if c, a_1 , a_2 , . . . , a_n \text{ are any real numbers, then } \]
\[c\left( a_1 + a_2 + . . . + a_n \right) = c a_1 + c a_2 + . . . + c a_n\]
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:
Given that tn+1 = 5tn + 4, t1 = 4, prove that tn = 5n − 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"`
Prove statement by using the Principle of Mathematical Induction for all n ∈ N, that:
2n + 1 < 2n, for all natual numbers n ≥ 3.
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:
32n – 1 is divisible by 8, 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.
Prove the statement by using the Principle of Mathematical Induction:
2 + 4 + 6 + ... + 2n = n2 + n 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.
Prove that for all n ∈ N.
cos α + cos(α + β) + cos(α + 2β) + ... + cos(α + (n – 1)β) = `(cos(alpha + ((n - 1)/2)beta)sin((nbeta)/2))/(sin beta/2)`.
Prove that, sinθ + sin2θ + sin3θ + ... + sinnθ = `((sin ntheta)/2 sin ((n + 1))/2 theta)/(sin theta/2)`, for all n ∈ N.
Prove that `1/(n + 1) + 1/(n + 2) + ... + 1/(2n) > 13/24`, for all natural numbers n > 1.
If P(n): 2n < n!, n ∈ N, then P(n) is true for all n ≥ ______.
