Advertisements
Advertisements
प्रश्न
12 + 32 + 52 + ... + (2n − 1)2 = \[\frac{1}{3}n(4 n^2 - 1)\]
Advertisements
उत्तर
Let P(n) be the given statement.
Now,
\[P(n) = 1^2 + 3^2 + 5^2 + . . . + (2n - 1 )^2 = \frac{1}{3}n(4 n^2 - 1)\]
\[\text{ Step 1:} \]
\[P(1) = 1^2 = 1 = \frac{1}{3} \times 1 \times (4 - 1)\]
\[\text{ Hence, P(1) is true } . \]
\[\text{ Step 2: } \]
\[\text{ Let P(m) be true } . \]
\[\text{ Then, } \]
\[ 1^2 + 3^2 + . . . + (2m - 1 )^2 = \frac{1}{3}m(4 m^2 - 1)\]
\[\text{ To prove: P(m + 1) is true whenever P(m) is true } . \]
\[\text{ That is, } \]
\[ 1^2 + 3^2 = . . . + (2m + 1 )^2 = \frac{1}{3}(m + 1)\left\{ 4(m + 1 )^2 - 1 \right\}\]
\[\text{ We know that P(m) is true } . \]
\[\text{ Thus, we have: } \]
\[ 1^2 + 3^2 + . . . + (2m - 1 )^2 = \frac{1}{3}m(4 m^2 - 1)\]
\[ \Rightarrow 1^2 + 3^2 + . . . + (2m - 1 )^2 + (2m + 1 )^2 = \frac{1}{3}m(4 m^2 - 1) + (2m + 1 )^2 \left[ \text{ Adding } (2m + 1 )^2 \text{ to both sides } \right]\]
\[ \Rightarrow P(m + 1) = \frac{1}{3}\left( 4 m^3 - m + 12 m^2 + 12m + 3 \right)\]
\[ \Rightarrow P(m + 1) = \frac{1}{3}(4 m^3 - m + 8 m^2 + 4m + 4 m^2 + 8m + 3)\]
\[ = \frac{1}{3}(m + 1)(4 m^2 + 8m + 3) \]
\[ = \frac{1}{3}(m + 1)(4(m + 1 )^2 - 1)\]
\[\text{ Thus, P(m + 1) is true } . \]
\[\text{ By the principle of mathematical induction, P(n) is true for all n } \in N . \]
APPEARS IN
संबंधित प्रश्न
Prove the following by using the principle of mathematical induction for all n ∈ N:
`1 + 3 + 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: 102n – 1 + 1 is divisible by 11
If P (n) is the statement "n3 + n is divisible by 3", prove that P (3) is true but P (4) is not true.
1 + 2 + 3 + ... + n = \[\frac{n(n + 1)}{2}\] i.e. the sum of the first n natural numbers is \[\frac{n(n + 1)}{2}\] .
1 + 3 + 5 + ... + (2n − 1) = n2 i.e., the sum of first n odd natural numbers is n2.
\[\frac{1}{1 . 4} + \frac{1}{4 . 7} + \frac{1}{7 . 10} + . . . + \frac{1}{(3n - 2)(3n + 1)} = \frac{n}{3n + 1}\]
2 + 5 + 8 + 11 + ... + (3n − 1) = \[\frac{1}{2}n(3n + 1)\]
1.2 + 2.3 + 3.4 + ... + n (n + 1) = \[\frac{n(n + 1)(n + 2)}{3}\]
a + (a + d) + (a + 2d) + ... (a + (n − 1) d) = \[\frac{n}{2}\left[ 2a + (n - 1)d \right]\]
11n+2 + 122n+1 is divisible by 133 for all n ∈ N.
\[\text{ A sequence } x_0 , x_1 , x_2 , x_3 , . . . \text{ is defined by letting } x_0 = 5 and x_k = 4 + x_{k - 1}\text{ for all natural number k . } \]
\[\text{ Show that } x_n = 5 + 4n \text{ for all n } \in N \text{ using mathematical induction .} \]
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 + 22 + 32 + .... + n2 = `("n"("n" + 1)(2"n" + 1))/6`
Prove by method of induction, for all n ∈ N:
1.3 + 3.5 + 5.7 + ..... to n terms = `"n"/3(4"n"^2 + 6"n" - 1)`
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)`
Answer the following:
Given that tn+1 = 5tn − 8, t1 = 3, prove by method of induction that tn = 5n−1 + 2
Prove statement by using the Principle of Mathematical Induction for all n ∈ N, that:
1 + 3 + 5 + ... + (2n – 1) = n2
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.
Let P(n): “2n < (1 × 2 × 3 × ... × n)”. Then the smallest positive integer for which P(n) is true is ______.
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:
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:
For any natural number n, xn – yn is divisible by x – y, where x and y are any integers with x ≠ y.
Prove the statement by using the Principle of Mathematical Induction:
n3 – n is divisible by 6, for each natural number n ≥ 2.
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.
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.
Prove that number of subsets of a set containing n distinct elements is 2n, for all n ∈ N.
If 10n + 3.4n+2 + k is divisible by 9 for all n ∈ N, then the least positive integral value of k is ______.
If P(n): 2n < n!, n ∈ N, then P(n) is true for all n ≥ ______.
Consider the statement: “P(n) : n2 – n + 41 is prime." Then which one of the following is true?
By using principle of mathematical induction for every natural number, (ab)n = ______.
