Advertisements
Advertisements
प्रश्न
Prove the statement by using the Principle of Mathematical Induction:
1 + 5 + 9 + ... + (4n – 3) = n(2n – 1) for all natural numbers n.
Advertisements
उत्तर
Let P(n) : 1 + 5 + 9 + … + (4n – 3) = n(2n – 1), for all natural numbers n.
P(1): 1 = 1(2 × 1 – 1) = 1, which is true.
Hence, P(1) is true.
Let us assume that P(n) is true for some natural number n = k.
∴ P(k) : 1 + 5 + 9 + … + (4k – 3) = k(2k – 1) .......(i)
Now, we have to prove that P(k + 1) is true.
P(k + 1) : 1 + 5 + 9 + … + (4k – 3) + [4(k + 1) – 3]
= 2k2 – k + 4k + 4 – 3
= 2k2 + 3k + 1
= (k + 1)( 2k + 1)
= (k + 1)[2(k + 1) – 1]
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n.
APPEARS IN
संबंधित प्रश्न
Prove the following by using the principle of mathematical induction for all n ∈ N: 1.2.3 + 2.3.4 + … + n(n + 1) (n + 2) = `(n(n+1)(n+2)(n+3))/(4(n+3))`
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+2+ 3+...+n<1/8(2n +1)^2`
Prove the following by using the principle of mathematical induction for all n ∈ N: x2n – y2n is divisible by x + y.
1 + 2 + 3 + ... + n = \[\frac{n(n + 1)}{2}\] i.e. the sum of the first n natural numbers is \[\frac{n(n + 1)}{2}\] .
\[\frac{1}{2 . 5} + \frac{1}{5 . 8} + \frac{1}{8 . 11} + . . . + \frac{1}{(3n - 1)(3n + 2)} = \frac{n}{6n + 4}\]
1.2 + 2.22 + 3.23 + ... + n.2n = (n − 1) 2n+1+2
1.3 + 2.4 + 3.5 + ... + n. (n + 2) = \[\frac{1}{6}n(n + 1)(2n + 7)\]
\[\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + . . . + \frac{1}{2^n} = 1 - \frac{1}{2^n}\]
a + ar + ar2 + ... + arn−1 = \[a\left( \frac{r^n - 1}{r - 1} \right), r \neq 1\]
52n −1 is divisible by 24 for all n ∈ N.
32n+7 is divisible by 8 for all n ∈ N.
32n+2 −8n − 9 is divisible by 8 for all n ∈ N.
x2n−1 + y2n−1 is divisible by x + y for all n ∈ 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:
`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:
`1/(3.5) + 1/(5.7) + 1/(7.9) + ...` to n terms = `"n"/(3(2"n" + 3))`
Prove by method of induction, for all n ∈ N:
(23n − 1) is divisible by 7
Answer the following:
Prove, by method of induction, for all n ∈ N
2 + 3.2 + 4.22 + ... + (n + 1)2n–1 = n.2n
Answer the following:
Prove by method of induction 52n − 22n is divisible by 3, for all n ∈ N
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
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:
n3 – 7n + 3 is divisible by 3, 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:
For any natural number n, 7n – 2n is divisible by 5.
Prove the statement by using the Principle of Mathematical Induction:
n3 – n is divisible by 6, for each natural number n ≥ 2.
Prove that `1/(n + 1) + 1/(n + 2) + ... + 1/(2n) > 13/24`, for all natural numbers n > 1.
If 10n + 3.4n+2 + k is divisible by 9 for all n ∈ N, then the least positive integral value of k is ______.
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 ≥ ______.
