Advertisements
Advertisements
प्रश्न
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))`
Advertisements
उत्तर
Let P(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))`, for all n ∈ N.
Step 1:
For n = 1, L.H.S. = `1/(3.4.5) = 1/60`
R.H.S. = `(1(1+1))/(6(1+3)(1+4))=2/(6(4)(5))=1/60`
∴ L.H.S. = R.H.S. for n = 1.
∴ P(1) is true.
Step 2:
Let us assume that for some k ∈ N, P(k) is true,
i.e., `1/(3.4.5) + 2/(4.5.6) + 3/(5.6.7) + ... + "k"/(("k" + 2)("k" + 3)("k" + 4)) = ("k"("k" + 1))/(6("k" + 3)("k" + 4))` ...(1)
Step 3:
To prove that P(k + 1) is true, i.e., to prove that
`1/(3.4.5) + 2/(4.5.6) + 3/(5.6.7) + ... + "k"/(("k" + 2)("k" + 3)("k" + 4)) + ("k" + 1)/(("k" + 3)("k" + 4)("k" + 5)) = (("k" + 1)("k" + 2))/(6("k" + 4)("k" + 5))`
Now, L.H.S. = `1/(3.4.5) + 2/(4.5.6) + 3/(5.6.7) + ... + "k"/(("k" + 2)("k" + 3)("k" + 4)) + ("k" + 1)/(("k" + 3)("k" + 4)("k" + 5))`
= `("k"("k" + 1))/(6("k" + 3)("k" + 4)) + ("k" + 1)/(("k" + 3)("k" + 4)("k" + 5))` ...[By (1)]
= `("k" + 1)/(("k" + 3)("k" + 4))["k"/6 + 1/("k" + 5)]`
= `("k" + 1)/(("k" + 3)("k" + 4))[("k"^2 + 5"k" + 6)/(6("k" + 5))]`
= `("k" + 1)/(("k" + 3)("k" + 4)) xx (("k" + 2)("k" + 3))/(6("k" + 5))`
= `(("k" + 1)("k" + 2))/(6("k" + 4)("k" + 5))`
= R.H.S.
∴ P(k + 1) is true.
Step 4:
From all the above steps and by the principle of mathematical induction P(n) is true for all n ∈ N,
i.e., `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))`, for all n ∈ 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:
(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: 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.
If P (n) is the statement "n2 − n + 41 is prime", prove that P (1), P (2) and P (3) are true. Prove also that P (41) is not 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.
1 + 3 + 32 + ... + 3n−1 = \[\frac{3^n - 1}{2}\]
\[\frac{1}{1 . 2} + \frac{1}{2 . 3} + \frac{1}{3 . 4} + . . . + \frac{1}{n(n + 1)} = \frac{n}{n + 1}\]
1.2 + 2.22 + 3.23 + ... + n.2n = (n − 1) 2n+1+2
2 + 5 + 8 + 11 + ... + (3n − 1) = \[\frac{1}{2}n(3n + 1)\]
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.
(ab)n = anbn for all n ∈ N.
11n+2 + 122n+1 is divisible by 133 for all n ∈ N.
Prove that n3 - 7n + 3 is divisible by 3 for all n \[\in\] N .
Prove that the number of subsets of a set containing n distinct elements is 2n, for all n \[\in\] N .
\[\text{ A sequence } a_1 , a_2 , a_3 , . . . \text{ is defined by letting } a_1 = 3 \text{ and } a_k = 7 a_{k - 1} \text{ for all natural numbers } k \geq 2 . \text{ Show that } a_n = 3 \cdot 7^{n - 1} \text{ for all } n \in N .\]
\[\text{ The distributive law from algebra states that for all real numbers} c, a_1 \text{ and } a_2 , \text{ we have } c\left( a_1 + a_2 \right) = c a_1 + c a_2 . \]
\[\text{ Use this law and mathematical induction to prove that, for all natural numbers, } n \geq 2, if c, a_1 , a_2 , . . . , a_n \text{ are any real numbers, then } \]
\[c\left( a_1 + a_2 + . . . + a_n \right) = c a_1 + c a_2 + . . . + c a_n\]
Prove by method of induction, for all n ∈ N:
3 + 7 + 11 + ..... + to n terms = n(2n+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:
13 + 33 + 53 + .... to n terms = n2(2n2 − 1)
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)`
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)`
Answer the following:
Prove, by method of induction, for all n ∈ N
2 + 3.2 + 4.22 + ... + (n + 1)2n–1 = n.2n
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. 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:
`sqrt(n) < 1/sqrt(1) + 1/sqrt(2) + ... + 1/sqrt(n)`, for all natural numbers n ≥ 2.
If 10n + 3.4n+2 + k is divisible by 9 for all n ∈ N, then the least positive integral value of k is ______.
If xn – 1 is divisible by x – k, then the least positive integral value of k is ______.
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.
