Advertisements
Advertisements
प्रश्न
\[\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + . . . + \frac{1}{2^n} = 1 - \frac{1}{2^n}\]
Advertisements
उत्तर
Let P(n) be the given statement.
Now,
\[P(n): \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + . . . + \frac{1}{2^n} = 1 - \frac{1}{2^n}\]
\[\text{ Step } 1: \]
\[ P(1) = \frac{1}{2} = 1 - \frac{1}{2^1}\]
\[\text{ Thus, P(1) is true .} \]
\[\text{ Step 2: } \]
\[\text{ Suppose P(m) is true .} \]
\[\text{ Then,} \]
\[\frac{1}{2} + \frac{1}{4} + . . . + \frac{1}{2^m} = 1 - \frac{1}{2^m}\]
\[\text{ To show: P(m + 1) is true whenever P(m) is true } . \]
\[\text{ That is, } \]
\[\frac{1}{2} + \frac{1}{4} + . . . + \frac{1}{2^{{}^{m + 1}}} = 1 - \frac{1}{2^{m + 1}}\]
\[\text{ Now, P(m) is true } . \]
\[\text{ Thus, we have: } \]
\[\frac{1}{2} + \frac{1}{4} + . . . + \frac{1}{2^m} = 1 - \frac{1}{2^m}\]
\[ \Rightarrow \frac{1}{2} + \frac{1}{4} + . . . + \frac{1}{2^m} + \frac{1}{2^{m + 1}} = 1 - \frac{1}{2^m} + \frac{1}{2^{m + 1}} \left[ \text{ Adding } \frac{1}{2^{m + 1}} \text{ to both sides } \right]\]
\[ \Rightarrow P(m + 1) = 1 - \frac{1}{2^m} + \frac{1}{2^m . 2} = 1 - \frac{1}{2^{{}^m}}\left( 1 - \frac{1}{2} \right) = 1 - \frac{1}{2^{m + 1}}\]
Thus , P ( m + 1) is true
$By the principle of mathematical induction, P(n) is true 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: 1.2.3 + 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: `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:
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: x2n – y2n is divisible by x + y.
Prove the following by using the principle of mathematical induction for all n ∈ N: 32n + 2 – 8n– 9 is divisible by 8.
If P (n) is the statement "n(n + 1) is even", then what is P(3)?
If P (n) is the statement "2n ≥ 3n" and if P (r) is true, prove that P (r + 1) is 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.
1 + 3 + 32 + ... + 3n−1 = \[\frac{3^n - 1}{2}\]
1 + 3 + 5 + ... + (2n − 1) = n2 i.e., the sum of first n odd natural numbers is n2.
\[\frac{1}{2 . 5} + \frac{1}{5 . 8} + \frac{1}{8 . 11} + . . . + \frac{1}{(3n - 1)(3n + 2)} = \frac{n}{6n + 4}\]
\[\frac{1}{1 . 4} + \frac{1}{4 . 7} + \frac{1}{7 . 10} + . . . + \frac{1}{(3n - 2)(3n + 1)} = \frac{n}{3n + 1}\]
\[\frac{1}{3 . 5} + \frac{1}{5 . 7} + \frac{1}{7 . 9} + . . . + \frac{1}{(2n + 1)(2n + 3)} = \frac{n}{3(2n + 3)}\]
1.3 + 2.4 + 3.5 + ... + n. (n + 2) = \[\frac{1}{6}n(n + 1)(2n + 7)\]
a + (a + d) + (a + 2d) + ... (a + (n − 1) d) = \[\frac{n}{2}\left[ 2a + (n - 1)d \right]\]
32n+7 is divisible by 8 for all n ∈ N.
2.7n + 3.5n − 5 is divisible by 24 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 \[\frac{a_n - \sqrt{A}}{a_n + \sqrt{A}} = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right) 2^{n - 1}\]
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))`
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
`[(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:
`(1 - 1/2^2).(1 - 1/3^2)...(1 - 1/n^2) = (n + 1)/(2n)`, for all natural numbers, n ≥ 2.
Prove by the Principle of Mathematical Induction that 1 × 1! + 2 × 2! + 3 × 3! + ... + n × n! = (n + 1)! – 1 for all natural numbers 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:
1 + 2 + 22 + ... + 2n = 2n+1 – 1 for all natural numbers n.
A sequence a1, a2, a3 ... is defined by letting a1 = 3 and ak = 7ak – 1 for all natural numbers k ≥ 2. Show that an = 3.7n–1 for all natural numbers.
A sequence d1, d2, d3 ... is defined by letting d1 = 2 and dk = `(d_(k - 1))/"k"` for all natural numbers, k ≥ 2. Show that dn = `2/(n!)` for all n ∈ N.
Prove that number of subsets of a set containing n distinct elements is 2n, for all n ∈ N.
For all n ∈ N, 3.52n+1 + 23n+1 is divisible by ______.
