Advertisements
Advertisements
प्रश्न
Prove by method of induction, for all n ∈ N:
2 + 4 + 6 + ..... + 2n = n (n+1)
Advertisements
उत्तर
Let P(n) ≡ 2 + 4 + 6 + …… + 2n = n(n + 1), for all n ∈ N
Step I:
Put n = 1
L.H.S. = 2
R.H.S. = 1(1 + 1) = 2 = L.H.S.
∴ P(n) is true for n = 1.
Step II:
Let us consider that P(n) is true for n = k
∴ 2 + 4 + 6 + ……. + 2k = k(k + 1) …(i)
Step III:
We have to prove that P(n) is true for n = k + 1 i.e., to prove that
2 + 4 + 6 + …. + 2(k + 1) = (k + 1) (k + 2)
L.H.S. = 2 + 4 + 6 + …… + 2 (k + 1)
= 2 + 4 + 6 + …… + 2k + 2(k + 1)
= k(k + 1) + 2(k + 1) …[From (i)]
= (k + 1).(k + 2)
= R.H.S.
∴ P(n) is true for n = k + 1
Step IV:
From all steps above by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 2 + 4 + 6 + …… + 2n = n (n + 1) for all n ∈ 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:
If P (n) is the statement "n(n + 1) is even", then what is P(3)?
\[\frac{1}{1 . 2} + \frac{1}{2 . 3} + \frac{1}{3 . 4} + . . . + \frac{1}{n(n + 1)} = \frac{n}{n + 1}\]
1 + 3 + 5 + ... + (2n − 1) = n2 i.e., the sum of first n odd natural numbers is n2.
\[\frac{1}{2 . 5} + \frac{1}{5 . 8} + \frac{1}{8 . 11} + . . . + \frac{1}{(3n - 1)(3n + 2)} = \frac{n}{6n + 4}\]
\[\frac{1}{3 . 7} + \frac{1}{7 . 11} + \frac{1}{11 . 5} + . . . + \frac{1}{(4n - 1)(4n + 3)} = \frac{n}{3(4n + 3)}\]
2 + 5 + 8 + 11 + ... + (3n − 1) = \[\frac{1}{2}n(3n + 1)\]
1.3 + 3.5 + 5.7 + ... + (2n − 1) (2n + 1) =\[\frac{n(4 n^2 + 6n - 1)}{3}\]
a + ar + ar2 + ... + arn−1 = \[a\left( \frac{r^n - 1}{r - 1} \right), r \neq 1\]
Prove that n3 - 7n + 3 is divisible by 3 for all n \[\in\] N .
Prove that 1 + 2 + 22 + ... + 2n = 2n+1 - 1 for all n \[\in\] N .
7 + 77 + 777 + ... + 777 \[{. . . . . . . . . . .}_{n - \text{ digits } } 7 = \frac{7}{81}( {10}^{n + 1} - 9n - 10)\]
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?
\[\text{ Given } a_1 = \frac{1}{2}\left( a_0 + \frac{A}{a_0} \right), a_2 = \frac{1}{2}\left( a_1 + \frac{A}{a_1} \right) \text{ and } a_{n + 1} = \frac{1}{2}\left( a_n + \frac{A}{a_n} \right) \text{ for } n \geq 2, \text{ where } a > 0, A > 0 . \]
\[\text{ Prove that } \frac{a_n - \sqrt{A}}{a_n + \sqrt{A}} = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right) 2^{n - 1} .\]
Prove by method of induction, for all n ∈ N:
12 + 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)`
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:
5 + 52 + 53 + .... + 5n = `5/4(5^"n" - 1)`
Prove by method of induction, for all n ∈ N:
(cos θ + i sin θ)n = cos (nθ) + i sin (nθ)
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 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:
22n – 1 is divisible by 3.
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.
Prove by induction that for all natural number n sinα + sin(α + β) + sin(α + 2β)+ ... + sin(α + (n – 1)β) = `(sin (alpha + (n - 1)/2 beta)sin((nbeta)/2))/(sin(beta/2))`
A student was asked to prove a statement P(n) by induction. He proved that P(k + 1) is true whenever P(k) is true for all k > 5 ∈ N and also that P(5) is true. On the basis of this he could conclude that P(n) is true ______.
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:
For any natural number n, xn – yn is divisible by x – y, where x and y are any integers with x ≠ y.
Show that `n^5/5 + n^3/3 + (7n)/15` is a natural number for all n ∈ N.
If xn – 1 is divisible by x – k, then the least positive integral value of k is ______.
If P(n): 2n < n!, n ∈ N, then P(n) is true for all n ≥ ______.
