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:
1.2 + 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 + 1/4 + 1/8 + ... + 1/2^n = 1 - 1/2^n`
Prove the following by using the principle of mathematical induction for all n ∈ N:
`(1+ 1/1)(1+ 1/2)(1+ 1/3)...(1+ 1/n) = (n + 1)`
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: 32n + 2 – 8n– 9 is divisible by 8.
If P (n) is the statement "n3 + n is divisible by 3", prove that P (3) is true but P (4) is not true.
If P (n) is the statement "2n ≥ 3n" and if P (r) is true, prove that P (r + 1) is true.
If P (n) is the statement "n2 + n is even", and if P (r) is true, then P (r + 1) is true.
12 + 22 + 32 + ... + n2 =\[\frac{n(n + 1)(2n + 1)}{6}\] .
\[\frac{1}{1 . 2} + \frac{1}{2 . 3} + \frac{1}{3 . 4} + . . . + \frac{1}{n(n + 1)} = \frac{n}{n + 1}\]
1 + 3 + 5 + ... + (2n − 1) = n2 i.e., the sum of first n odd natural numbers is n2.
1.3 + 2.4 + 3.5 + ... + n. (n + 2) = \[\frac{1}{6}n(n + 1)(2n + 7)\]
a + ar + ar2 + ... + arn−1 = \[a\left( \frac{r^n - 1}{r - 1} \right), r \neq 1\]
52n −1 is divisible by 24 for all n ∈ N.
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.
7 + 77 + 777 + ... + 777 \[{. . . . . . . . . . .}_{n - \text{ digits } } 7 = \frac{7}{81}( {10}^{n + 1} - 9n - 10)\]
\[\text{ A sequence } a_1 , a_2 , a_3 , . . . \text{ is defined by letting } a_1 = 3 \text{ and } a_k = 7 a_{k - 1} \text{ for all natural numbers } k \geq 2 . \text{ Show that } a_n = 3 \cdot 7^{n - 1} \text{ for all } n \in N .\]
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)
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))`
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.
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"):}`
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:
32n – 1 is divisible by 8, for all natural numbers n.
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:
n2 < 2n for all natural numbers n ≥ 5.
Prove the statement by using the Principle of Mathematical Induction:
2 + 4 + 6 + ... + 2n = n2 + n for all natural numbers 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 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.
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)`.
Show that `n^5/5 + n^3/3 + (7n)/15` is a natural number for all n ∈ N.
If xn – 1 is divisible by x – k, then the least positive integral value of k is ______.
Consider the statement: “P(n) : n2 – n + 41 is prime." Then which one of the following is true?
