Advertisements
Advertisements
Question
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
Solution
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
RELATED QUESTIONS
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: 41n – 14n is a multiple of 27.
If P (n) is the statement "2n ≥ 3n" and if P (r) is true, prove that P (r + 1) is true.
\[\frac{1}{1 . 2} + \frac{1}{2 . 3} + \frac{1}{3 . 4} + . . . + \frac{1}{n(n + 1)} = \frac{n}{n + 1}\]
\[\frac{1}{2 . 5} + \frac{1}{5 . 8} + \frac{1}{8 . 11} + . . . + \frac{1}{(3n - 1)(3n + 2)} = \frac{n}{6n + 4}\]
\[\frac{1}{3 . 5} + \frac{1}{5 . 7} + \frac{1}{7 . 9} + . . . + \frac{1}{(2n + 1)(2n + 3)} = \frac{n}{3(2n + 3)}\]
1.3 + 2.4 + 3.5 + ... + n. (n + 2) = \[\frac{1}{6}n(n + 1)(2n + 7)\]
1.3 + 3.5 + 5.7 + ... + (2n − 1) (2n + 1) =\[\frac{n(4 n^2 + 6n - 1)}{3}\]
a + ar + ar2 + ... + arn−1 = \[a\left( \frac{r^n - 1}{r - 1} \right), r \neq 1\]
52n+2 −24n −25 is divisible by 576 for all n ∈ N.
72n + 23n−3. 3n−1 is divisible by 25 for all n ∈ N.
11n+2 + 122n+1 is divisible by 133 for all n ∈ N.
x2n−1 + y2n−1 is divisible by x + y for all n ∈ N.
Prove by method of induction, for all n ∈ N:
(23n − 1) is divisible by 7
Prove by method of induction, for all n ∈ N:
(24n−1) is divisible by 15
Prove by method of induction, for all n ∈ N:
(cos θ + i sin θ)n = cos (nθ) + i sin (nθ)
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:
`sum_(t = 1)^(n - 1) t(t + 1) = (n(n - 1)(n + 1))/3`, for all natural numbers n ≥ 2.
Prove statement by using the Principle of Mathematical Induction for all n ∈ N, that:
22n – 1 is divisible by 3.
Define the sequence a1, a2, a3 ... as follows:
a1 = 2, an = 5 an–1, for all natural numbers n ≥ 2.
Use the Principle of Mathematical Induction to show that the terms of the sequence satisfy the formula an = 2.5n–1 for all natural numbers.
Show by the Principle of Mathematical Induction that the sum Sn of the n term of the series 12 + 2 × 22 + 32 + 2 × 42 + 52 + 2 × 62 ... is given by
Sn = `{{:((n(n + 1)^2)/2",", "if n is even"),((n^2(n + 1))/2",", "if n is odd"):}`
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. Justify your answer.
Prove the statement by using the Principle of Mathematical Induction:
4n – 1 is divisible by 3, for each natural number n.
Prove the statement by using the Principle of Mathematical Induction:
23n – 1 is divisible by 7, for all natural numbers n.
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.
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.
For all n ∈ N, 3.52n+1 + 23n+1 is divisible by ______.
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.
Consider the statement: “P(n) : n2 – n + 41 is prime." Then which one of the following is true?
