Advertisements
Advertisements
प्रश्न
Prove by method of induction, for all n ∈ N:
12 + 22 + 32 + .... + n2 = `("n"("n" + 1)(2"n" + 1))/6`
Advertisements
उत्तर
Let P(n) ≡ 12 + 22 + 32 + .... + n2 = `("n"("n" + 1)(2"n" + 1))/6`, for all n ∈ N
Step I:
Put n = 1
L.H.S. = 12 = 1
R.H.S. = `(1(1 + 1)[2(1) + 1])/6 = 6/6` = 1
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1
Step II:
Let us consider that P(n) is true for n = k
∴ 12 + 22 + 32 + .... + k2 = `("k"("k" + 1)(2"k" + 1))/6` ...(i)
Step III:
We have to prove that P(n) is true for n = k + 1
i.e., to prove that
12 + 22 + 32 + …. + (k + 1)2
= `(("k" + 1)[("k" + 1) + 1][2("k" + 1) + 1])/6`
= `(("k" + 1)("k" + 2)(2"k" + 3))/6`
L.H.S. = 12 + 22 + 32 + …. + (k + 1)2
= 12 + 22 + 32 + …. + k2 + (k + 1)2
= `("k"("k" + 1)(2"k" + 1))/6 + ("k" + 1)^2` ...[From (i)]
= `("k" + 1)[("k"(2"k" + 1))/6 + ("k" + 1)]`
= `("k" + 1)[(2"k"^2 + "k" + 6"k" + 6)/6]`
= `(("k" + 1)(2"k"^2 + 7"k" + 6))/6`
= `(("k" + 1)("k" + 2)(2"k" + 3))/6`
= 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.
∴ 12 + 22 + 32 + .... + n2 = `("n"("n" + 1)(2"n" + 1))/6` for all n ∈ 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: `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:
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: 41n – 14n is a multiple of 27.
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.
\[\frac{1}{1 . 4} + \frac{1}{4 . 7} + \frac{1}{7 . 10} + . . . + \frac{1}{(3n - 2)(3n + 1)} = \frac{n}{3n + 1}\]
\[\frac{1}{3 . 7} + \frac{1}{7 . 11} + \frac{1}{11 . 5} + . . . + \frac{1}{(4n - 1)(4n + 3)} = \frac{n}{3(4n + 3)}\]
1.2 + 2.22 + 3.23 + ... + n.2n = (n − 1) 2n+1+2
2 + 5 + 8 + 11 + ... + (3n − 1) = \[\frac{1}{2}n(3n + 1)\]
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) =\[\frac{n(4 n^2 + 6n - 1)}{3}\]
52n −1 is divisible by 24 for all n ∈ N.
52n+2 −24n −25 is divisible by 576 for all n ∈ N.
Prove that 1 + 2 + 22 + ... + 2n = 2n+1 - 1 for all n \[\in\] N .
\[\text{ Given } a_1 = \frac{1}{2}\left( a_0 + \frac{A}{a_0} \right), a_2 = \frac{1}{2}\left( a_1 + \frac{A}{a_1} \right) \text{ and } a_{n + 1} = \frac{1}{2}\left( a_n + \frac{A}{a_n} \right) \text{ for } n \geq 2, \text{ where } a > 0, A > 0 . \]
\[\text{ Prove that } \frac{a_n - \sqrt{A}}{a_n + \sqrt{A}} = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right) 2^{n - 1} .\]
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 \[S_n = \binom{\frac{n \left( n + 1 \right)^2}{2}, \text{ if n is even} }{\frac{n^2 \left( n + 1 \right)}{2}, \text{ if n is odd } }\]
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:
(24n−1) is divisible by 15
Prove by method of induction, for all n ∈ N:
`[(1, 2),(0, 1)]^"n" = [(1, 2"n"),(0, 1)]` ∀ n ∈ N
Answer the following:
Given that tn+1 = 5tn − 8, t1 = 3, prove by method of induction that tn = 5n−1 + 2
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 loga xn = n logax, x > 0, n ∈ N
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
The distributive law from algebra says that for all real numbers c, a1 and a2, we have c(a1 + a2) = ca1 + ca2.
Use this law and mathematical induction to prove that, for all natural numbers, n ≥ 2, if c, a1, a2, ..., an are any real numbers, then c(a1 + a2 + ... + an) = ca1 + ca2 + ... + can.
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:
n3 – n is divisible by 6, for each natural number n ≥ 2.
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:
1 + 2 + 22 + ... + 2n = 2n+1 – 1 for all natural numbers n.
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)`.
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.
Consider the statement: “P(n) : n2 – n + 41 is prime." Then which one of the following is true?
