Advertisements
Advertisements
Question
Solution
Let P(n) be the given statement.
Now,
\[P(n): \frac{n^7}{7} + \frac{n^5}{5} + \frac{n^3}{3} + \frac{n^2}{2} - \frac{37}{210}n \text{ is a positive integer .} \]
\[\text{ Step 1:} \]
\[P(1) = \frac{1}{7} + \frac{1}{5} + \frac{1}{3} + \frac{1}{2} - \frac{37}{210} = \frac{30 + 42 + 70 + 105 - 37}{210} = \frac{210}{210} = 1 \]
\[\text{ It is a positive integer .} \]
\[\text{ Thus, P(1) is true } . \]
\[\text{ Step } 2: \]
\[\text{ Let P(m) be true } . \]
\[\text{ Then } , \frac{m^7}{7} + \frac{m^5}{5} + \frac{m^3}{3} + \frac{m^2}{2} - \frac{37}{210}m \text{ is a positive integer } . \]
\[Let \frac{m^7}{7} + \frac{m^5}{5} + \frac{m^3}{3} + \frac{m^2}{2} - \frac{37}{210}m = \lambda \text{ for some } \lambda \in \text{ positive } N . \]
\[\text{ To show: } P\left( m + 1 \right) \text { is a positive integer } . \]
\[\text{ Now } , \]
\[P(m + 1) = \frac{\left( m + 1 \right)^7}{7} + \frac{\left( m + 1 \right)^5}{5} + \frac{\left( m + 1 \right)^3}{3} + \frac{\left( m + 1 \right)^2}{2} - \frac{37}{210}\left( m + 1 \right)\]
\[ = \frac{1}{7}\left( m^7 + 7 m^6 + 21 m^5 + 35 m^4 + 35 m^3 + 21 m^2 + 7m + 1 \right)\]
\[ + \frac{1}{5}\left( m^5 + 5 m^4 + 10 m^3 + 10 m^2 + 5m + 1 \right) + \frac{1}{3}\left( m^3 + 3 m^2 + 3m + 1 \right) + \frac{1}{2}\left( m^2 + 2m + 1 \right) - \frac{37}{210}m - \frac{37}{210} \]
\[ = \left[ \frac{m^7}{7} + \frac{m^5}{5} + \frac{m^3}{3} + \frac{m^2}{2} - \frac{37}{210}m \right] + m^6 + 3 m^5 + 6 m^4 + 7 m^3 + 6 m^2 + 4m\]
\[ = \lambda + m^6 + 3 m^5 + 6 m^4 + 7 m^3 + 6 m^2 + 4m\]
\[\text{ It is a positive integer, as \lambda is a positive integer } . \]
\[\text{ Thus } , P\left( m + 1 \right) \text{ is true } , \]
\[\text{ By the principle of mathematical induction, P(n) is true for all } n \in N . \]
APPEARS IN
RELATED QUESTIONS
Prove the following by using the principle of mathematical induction for all n ∈ N:
`1^3 + 2^3 + 3^3 + ... + n^3 = ((n(n+1))/2)^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 + 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+ 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:
`1/1.4 + 1/4.7 + 1/7.10 + ... + 1/((3n - 2)(3n + 1)) = n/((3n + 1))`
Prove the following by using the principle of mathematical induction for all n ∈ N: 32n + 2 – 8n– 9 is divisible by 8.
Prove the following by using the principle of mathematical induction for all n ∈ N: 41n – 14n is a multiple of 27.
Prove the following by using the principle of mathematical induction for all n ∈ N (2n +7) < (n + 3)2
Given an example of a statement P (n) such that it is true for all n ∈ N.
1 + 3 + 32 + ... + 3n−1 = \[\frac{3^n - 1}{2}\]
1.2 + 2.22 + 3.23 + ... + n.2n = (n − 1) 2n+1+2
\[\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + . . . + \frac{1}{2^n} = 1 - \frac{1}{2^n}\]
12 + 32 + 52 + ... + (2n − 1)2 = \[\frac{1}{3}n(4 n^2 - 1)\]
52n+2 −24n −25 is divisible by 576 for all n ∈ N.
n(n + 1) (n + 5) is a multiple of 3 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 \[\frac{a_n - \sqrt{A}}{a_n + \sqrt{A}} = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right) 2^{n - 1}\]
\[\frac{(2n)!}{2^{2n} (n! )^2} \leq \frac{1}{\sqrt{3n + 1}}\] for all n ∈ N .
\[\text{ The distributive law from algebra states that for all real numbers} c, a_1 \text{ and } a_2 , \text{ we have } c\left( a_1 + a_2 \right) = c a_1 + c a_2 . \]
\[\text{ Use this law and mathematical induction to prove that, for all natural numbers, } n \geq 2, if c, a_1 , a_2 , . . . , a_n \text{ are any real numbers, then } \]
\[c\left( a_1 + a_2 + . . . + a_n \right) = c a_1 + c a_2 + . . . + c a_n\]
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:
(23n − 1) is divisible by 7
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:
`(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.
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`
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:
n3 – 7n + 3 is divisible by 3, for all natural numbers n.
Prove the statement by using the Principle of Mathematical Induction:
2 + 4 + 6 + ... + 2n = n2 + n 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 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, sinθ + sin2θ + sin3θ + ... + sinnθ = `((sin ntheta)/2 sin ((n + 1))/2 theta)/(sin theta/2)`, for all n ∈ N.
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 ______.
