Advertisements
Advertisements
प्रश्न
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))`
Advertisements
उत्तर
Let P(n) ≡ `1/(3.5) + 1/(5.7) + 1/(7.9) + ...` to n terms = `"n"/(3(2"n" + 3))`, for all n ∈ N
But the first factor in each term of the denominator
i.e., 3, 5, 7, … are in A.P. with a = 3 and d = 2.
∴ nth term = a + (n – 1)d = 3 + (n – 1)2 = (2n + 1)
Also, second factor in each term of denominator
i.e., 5, 7, 9, … are in A.P. with a = 5 and d = 2.
∴ nth term = a + (n – 1)d = 5 + (n – 1)2 = (2n + 3)
∴ nth term tn = `1/((2"n" + 1)(2"n" + 3))`
∴ P(n) ≡ `1/(3.5) + 1/(5.7) + 1/(7.9) + ... + 1/((2"n" + 1)(2"n" + 3)) = "n"/(3(2"n" + 3))`
Step I:
Put n = 1
L.H.S. = `1/(3.5) = 1/15`
R.H.S. = `1/(3[2(1) + 3]) = 1/(3(2 + 3)) = 1/15` = L.H.S.
∴ P(n) is true for n = 1
Step II:
Let us consider that P(n) is true for n = k
∴ `1/(3.5) + 1/(5.7) + 1/(7.9) + ... + 1/((2"k" + 1).(2"k" + 3)) = "k"/(3(2"k" + 3))` ...(i)
Step III:
We have to prove that P(n) is true for n = k + 1
i.e., to prove that
`1/(3.5) + 1/(5.7) + 1/(7.9) + ... + 1/((2"k" + 3).(2"k" + 5)) = ("k" + 1)/(3(2"k" + 5))`
L.H.S. = `1/(3.5) + 1/(5.7) + 1/(7.9) + ... + 1/((2"k" + 3)(2"k" + 5))`
= `1/(3.5) + 1/(5.7) + 1/(7.9) + ... + 1/((2"k" + 1)(2"k" + 3)) + 1/((2"k" + 3)(2"k" + 5))`
= `"k"/(3(2"k" + 3)) + 1/((2"k" + 3)(2"k" + 5))` ...[From (i)]
= `("k"(2"k" + 5) + 3)/(3(2"k" + 3)(2"k" + 5))`
= `(2"k"^2 + 5"k" + 3)/(3(2"k" + 3)(2"k" + 5))`
= `(2"k"^2 + 2"k" + 3"k" + 3)/(3(2"k" + 3)(2"k" + 5)`
= `(2"k"("k" + 1) + 3("k" + 1))/(3(2"k" + 3)(2"k" + 5))`
= `((2"k" + 3)("k" + 1))/(3("2k" + 3)(2"k" + 5))`
= `("k" + 1)/(3(2"k" + 5))`
= R.H.S.
∴ P(n) is true for n = k + 1
Step IV:
From all steps above by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ `1/(3.5) + 1/(5.7) + 1/(7.9) + ...` to n terms = `"n"/(3(2"n" + 3))` for all n ∈ 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: 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:
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: 1.2 + 2.22 + 3.22 + … + n.2n = (n – 1) 2n+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:
Prove the following by using the principle of mathematical induction for all n ∈ N: n (n + 1) (n + 5) is a multiple of 3.
Prove the following by using the principle of mathematical induction for all n ∈ N: 102n – 1 + 1 is divisible by 11
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: 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
If P (n) is the statement "n2 + n is even", and if P (r) is true, then P (r + 1) is true.
\[\frac{1}{3 . 7} + \frac{1}{7 . 11} + \frac{1}{11 . 5} + . . . + \frac{1}{(4n - 1)(4n + 3)} = \frac{n}{3(4n + 3)}\]
1.2 + 2.22 + 3.23 + ... + n.2n = (n − 1) 2n+1+2
2 + 5 + 8 + 11 + ... + (3n − 1) = \[\frac{1}{2}n(3n + 1)\]
1.3 + 2.4 + 3.5 + ... + n. (n + 2) = \[\frac{1}{6}n(n + 1)(2n + 7)\]
1.3 + 3.5 + 5.7 + ... + (2n − 1) (2n + 1) =\[\frac{n(4 n^2 + 6n - 1)}{3}\]
a + (a + d) + (a + 2d) + ... (a + (n − 1) d) = \[\frac{n}{2}\left[ 2a + (n - 1)d \right]\]
52n −1 is divisible by 24 for all n ∈ N.
\[\text{ Given } a_1 = \frac{1}{2}\left( a_0 + \frac{A}{a_0} \right), a_2 = \frac{1}{2}\left( a_1 + \frac{A}{a_1} \right) \text{ and } a_{n + 1} = \frac{1}{2}\left( a_n + \frac{A}{a_n} \right) \text{ for } n \geq 2, \text{ where } a > 0, A > 0 . \]
\[\text{ Prove that } \frac{a_n - \sqrt{A}}{a_n + \sqrt{A}} = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right) 2^{n - 1} .\]
\[\text{ Let } P\left( n \right) \text{ be the statement } : 2^n \geq 3n . \text{ If } P\left( r \right) \text{ is true, then show that } P\left( r + 1 \right) \text{ is true . Do you conclude that } P\left( n \right)\text{ is true for all n } \in N?\]
Show by the Principle of Mathematical induction that the sum Sn of then terms of the series \[1^2 + 2 \times 2^2 + 3^2 + 2 \times 4^2 + 5^2 + 2 \times 6^2 + 7^2 + . . .\] is given by \[S_n = \binom{\frac{n \left( n + 1 \right)^2}{2}, \text{ if n is even} }{\frac{n^2 \left( n + 1 \right)}{2}, \text{ if n is odd } }\]
Prove that the number of subsets of a set containing n distinct elements is 2n, for all n \[\in\] 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:
12 + 22 + 32 + .... + n2 = `("n"("n" + 1)(2"n" + 1))/6`
Prove by method of induction, for all n ∈ N:
5 + 52 + 53 + .... + 5n = `5/4(5^"n" - 1)`
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
`1/(3.4.5) + 2/(4.5.6) + 3/(5.6.7) + ... + "n"/(("n" + 2)("n" + 3)("n" + 4)) = ("n"("n" + 1))/(6("n" + 3)("n" + 4))`
Answer the following:
Prove by method of induction 152n–1 + 1 is divisible by 16, for all n ∈ N.
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.
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`
A sequence b0, b1, b2 ... is defined by letting b0 = 5 and bk = 4 + bk – 1 for all natural numbers k. Show that bn = 5 + 4n for all natural number n using mathematical induction.
Prove that for all n ∈ N.
cos α + cos(α + β) + cos(α + 2β) + ... + cos(α + (n – 1)β) = `(cos(alpha + ((n - 1)/2)beta)sin((nbeta)/2))/(sin beta/2)`.
