Advertisements
Advertisements
प्रश्न
1.3 + 3.5 + 5.7 + ... + (2n − 1) (2n + 1) =\[\frac{n(4 n^2 + 6n - 1)}{3}\]
Advertisements
उत्तर
Let P(n) be the given statement.
Now,
\[P(n) = 1 . 3 + 3 . 5 + 5 . 7 + . . . + (2n - 1)(2n + 1) = \frac{n(4 n^2 + 6n - 1)}{3}\]
\[\text{ Step } 1: \]
\[P(1) = 1 . 3 = 3 = \frac{1(4 \times \left( 1 \right)^2 + 6 \times 1 - 1)}{3}\]
\[\text{ Hence, P(1) is true } . \]
\[\text{ Step 2: } \]
\[\text{ Let P(m) be true} . \]
\[\text{ Then,} \]
\[1 . 3 + 3 . 5 + . . . + (2m - 1)(2m + 1) = \frac{m(4 m^2 + 6m - 1)}{3}\]
\[\text{ To prove: P(m + 1) is true} . \]
\[\text{ That is, } \]
\[1 . 3 + 3 . 5 + . . . + (2m + 1)(2m + 3) = \frac{(m + 1)\left[ 4(m + 1 )^2 + 6\left( m + 1 \right) - 1 \right]}{3}\]
\[ \text{ Now, P(m) is equal to: } \]
\[1 . 3 + 3 . 5 + . . . + (2m - 1)(2m + 1) = \frac{m(4 m^2 + 6m - 1)}{3}\]
\[ \Rightarrow 1 . 3 + 3 . 5 + . . . + (2m - 1)(2m + 1) + (2m + 1)(2m + 3) = \frac{m(4 m^2 + 6m - 1)}{3} + (2m + 1)(2m + 3) \left[ \text{ Adding } (2m + 1)(2m + 3) \text{ to both sides } \right]\]
\[ \Rightarrow P(m + 1) = \frac{m(4 m^2 + 6m - 1) + 3(4 m^2 + 8m + 3)}{3}\]
\[ \Rightarrow P(m + 1) = \frac{4 m^3 + 6 m^2 - m + 12 m^2 + 24m + 9}{3} = \frac{4 m^3 + 18 m^2 + 23m + 9}{3}\]
\[ \Rightarrow P(m + 1) = \frac{4m( m^2 + 2m + 1) + 10 m^2 + 19m + 9}{3}\]
\[ = \frac{4m(m + 1 )^2 + (10m + 9)(m + 1)}{3}\]
\[ = \frac{(m + 1)\left[ 4m(m + 1) + 10m + 9 \right]}{3}\]
\[ = \frac{(m + 1)}{3}(4 m^2 + 8m + 4 + 6m + 5)\]
\[ = \frac{(m + 1)\left[ 4(m + 1 )^2 + 6\left( m + 1 \right) - 1 \right]}{3}\]
\[\text{ Thus, P(m + 1) is true .} \]
\[\text{ By the principle of mathematical induction, P(n) is true for all n} \in N .\]
APPEARS IN
संबंधित प्रश्न
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: 1.2.3 + 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+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:
`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: `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 "n(n + 1) is even", then what is P(3)?
Given an example of a statement P (n) such that it is true for all n ∈ N.
2.7n + 3.5n − 5 is divisible by 24 for all n ∈ N.
Let P(n) be the statement : 2n ≥ 3n. If P(r) is true, show that P(r + 1) is true. Do you conclude that P(n) is true for all n ∈ N?
x2n−1 + y2n−1 is divisible by x + y for all n ∈ N.
\[\text{ Given } a_1 = \frac{1}{2}\left( a_0 + \frac{A}{a_0} \right), a_2 = \frac{1}{2}\left( a_1 + \frac{A}{a_1} \right) \text{ and } a_{n + 1} = \frac{1}{2}\left( a_n + \frac{A}{a_n} \right) \text{ for } n \geq 2, \text{ where } a > 0, A > 0 . \]
\[\text{ Prove that } \frac{a_n - \sqrt{A}}{a_n + \sqrt{A}} = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right) 2^{n - 1} .\]
Prove by method of induction, for all n ∈ N:
2 + 4 + 6 + ..... + 2n = n (n+1)
Prove by method of induction, for all n ∈ N:
12 + 32 + 52 + .... + (2n − 1)2 = `"n"/3 (2"n" − 1)(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:
3n − 2n − 1 is divisible by 4
Prove by method of induction, for all n ∈ N:
(cos θ + i sin θ)n = cos (nθ) + i sin (nθ)
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
12 + 42 + 72 + ... + (3n − 2)2 = `"n"/2 (6"n"^2 - 3"n" - 1)`
Prove statement by using the Principle of Mathematical Induction for all n ∈ N, that:
2n + 1 < 2n, for all natual numbers n ≥ 3.
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.
Prove by the Principle of Mathematical Induction that 1 × 1! + 2 × 2! + 3 × 3! + ... + n × n! = (n + 1)! – 1 for all natural numbers n.
A student was asked to prove a statement P(n) by induction. He proved that P(k + 1) is true whenever P(k) is true for all k > 5 ∈ N and also that P(5) is true. On the basis of this he could conclude that P(n) 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
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:
For any natural number n, 7n – 2n is divisible by 5.
Prove the statement by using the Principle of Mathematical Induction:
n(n2 + 5) is divisible by 6, for each natural number n.
Prove the statement by using the Principle of Mathematical Induction:
2n < (n + 2)! for all natural number 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.
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.
