Advertisements
Advertisements
Question
Prove that 1 + 2 + 22 + ... + 2n = 2n+1 - 1 for all n \[\in\] N .
Advertisements
Solution
\[Let p\left( n \right): 1 + 2 + 2^2 + . . . + 2^n = 2^{n + 1} - 1 \forall n \in N\]
\[\text{ Step I: For } n = 1, \]
\[LHS = 1 + 2^1 = 3\]
\[RHS = 2^{1 + 1} - 1 = 2^2 - 1 = 4 - 1 = 3\]
\[As, LHS = RHS\]
\[\text{ So, it is true for n } = 1 . \]
\[\text{ Step II: For n } = k, \]
\[\text{ Let } p\left( k \right): 1 + 2 + 2^2 + . . . + 2^k = 2^{k + 1} - 1\text{ be true } \forall k \in N\]
\[\text{ Step III: For } n = k + 1, \]
\[LHS = 1 + 2 + 2^2 + . . . + 2^k + 2^{k + 1} \]
\[ = 2^{k + 1} - 1 + 2^{k + 1} \left(\text{ Using step } II \right)\]
\[ = 2 \times 2^{k + 1} - 1\]
\[ = 2^{k + 1 + 1} - 1\]
\[ = 2^{k + 2} - 1\]
\[RHS = 2^\left( k + 1 \right) + 1 - 1 = 2^{k + 2} - 1\]
\[As, LHS = RHS\]
\[\text{ So, it is also true for n } = k + 1 .\]
Hence, 1 + 2 + 22 + ... + 2n = 2n+1 - 1 for all n \[\in\] N .
APPEARS IN
RELATED QUESTIONS
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 + 2.3 + 3.4+ ... + n(n+1) = `[(n(n+1)(n+2))/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:
`(1+ 1/1)(1+ 1/2)(1+ 1/3)...(1+ 1/n) = (n + 1)`
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: 41n – 14n is a multiple of 27.
If P (n) is the statement "n2 + n is even", and if P (r) is true, then P (r + 1) is true.
\[\frac{1}{2 . 5} + \frac{1}{5 . 8} + \frac{1}{8 . 11} + . . . + \frac{1}{(3n - 1)(3n + 2)} = \frac{n}{6n + 4}\]
\[\frac{1}{3 . 5} + \frac{1}{5 . 7} + \frac{1}{7 . 9} + . . . + \frac{1}{(2n + 1)(2n + 3)} = \frac{n}{3(2n + 3)}\]
1.3 + 2.4 + 3.5 + ... + n. (n + 2) = \[\frac{1}{6}n(n + 1)(2n + 7)\]
a + ar + ar2 + ... + arn−1 = \[a\left( \frac{r^n - 1}{r - 1} \right), r \neq 1\]
32n+2 −8n − 9 is divisible by 8 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 \[\frac{a_n - \sqrt{A}}{a_n + \sqrt{A}} = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right) 2^{n - 1}\]
x2n−1 + y2n−1 is divisible by x + y for all n ∈ N.
\[\text{ Let } P\left( n \right) \text{ be the statement } : 2^n \geq 3n . \text{ If } P\left( r \right) \text{ is true, then show that } P\left( r + 1 \right) \text{ is true . Do you conclude that } P\left( n \right)\text{ is true for all n } \in N?\]
Prove that the number of subsets of a set containing n distinct elements is 2n, 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:
1.2 + 2.3 + 3.4 + ..... + n(n + 1) = `"n"/3 ("n" + 1)("n" + 2)`
Prove by method of induction, for all n ∈ N:
(24n−1) is divisible by 15
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, for all n ∈ N
8 + 17 + 26 + … + (9n – 1) = `"n"/2(9"n" + 7)`
Answer the following:
Prove by method of induction loga xn = n logax, x > 0, 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.
Prove by the Principle of Mathematical Induction that 1 × 1! + 2 × 2! + 3 × 3! + ... + n × n! = (n + 1)! – 1 for all natural numbers n.
Show by the Principle of Mathematical Induction that the sum Sn of the n term of the series 12 + 2 × 22 + 32 + 2 × 42 + 52 + 2 × 62 ... is given by
Sn = `{{:((n(n + 1)^2)/2",", "if n is even"),((n^2(n + 1))/2",", "if n is odd"):}`
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:
4n – 1 is divisible by 3, for each natural number n.
Prove the statement by using the Principle of Mathematical Induction:
2n < (n + 2)! for all natural number n.
Prove that number of subsets of a set containing n distinct elements is 2n, for all n ∈ N.
If P(n): 2n < n!, n ∈ N, then P(n) is true for all n ≥ ______.
By using principle of mathematical induction for every natural number, (ab)n = ______.
