Advertisements
Advertisements
Question
2 + 5 + 8 + 11 + ... + (3n − 1) = \[\frac{1}{2}n(3n + 1)\]
Advertisements
Solution
Let P(n) be the given statement.
Now,
\[P(n) = 2 + 5 + 8 + . . . + (3n - 1) = \frac{1}{2}n(3n + 1)\]
\[\text{ Step} 1: \]
\[P(1) = 2 = \frac{1}{2} \times 1(3 + 1) \]
\[\text{ Hence, P(1) is true .} \]
\[\text{ Step} 2: \]
\[\text{ Let P(m) be true } . \]
\[\text{ Then, } \]
\[2 + 5 + 8 + . . . + (3m - 1) = \frac{1}{2}m(3m + 1)\]
\[\text{ To prove: P(m + 1) is true } . \]
\[\text{ That is, } \]
\[2 + 5 + 8 + . . . + (3m + 2) = \frac{1}{2}(m + 1)(3m + 4)\]
\[\text{ P(m) is equal to:} \]
\[2 + 5 + 8 + . . . + (3m - 1) = \frac{1}{2}m(3m + 1)\]
\[\text{ Thus, we have:} \]
\[2 + 5 + 8 + . . . + (3m - 1) + (3m + 2) = \frac{1}{2}m(3m + 1) + (3m + 2) \left[ \text{ Adding } (3m + 2) \text{ to both sides } ] \right]\]
\[ \Rightarrow 2 + 5 + 8 + . . . + (3m + 2) = \frac{1}{2}(3 m^2 + m + 6m + 4) = \frac{1}{2}(3 m^2 + 7m + 4)\]
\[ \Rightarrow 2 + 5 + 8 + . . . + (3m + 2) = \frac{1}{2}(3m + 4)(m + 1)\]
\[\text{ Thus, P(m + 1) is true } . \]
\[\text{ By the principle of mathematical induction, P(n) is true for all n} \in N .\]
APPEARS IN
RELATED QUESTIONS
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:
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: `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: n (n + 1) (n + 5) is a multiple of 3.
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 "n(n + 1) is even", then what is P(3)?
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 + 41 is prime", prove that P (1), P (2) and P (3) are true. Prove also that P (41) is not true.
1 + 2 + 3 + ... + n = \[\frac{n(n + 1)}{2}\] i.e. the sum of the first n natural numbers is \[\frac{n(n + 1)}{2}\] .
12 + 22 + 32 + ... + n2 =\[\frac{n(n + 1)(2n + 1)}{6}\] .
1 + 3 + 5 + ... + (2n − 1) = n2 i.e., the sum of first n odd natural numbers is n2.
1.2 + 2.22 + 3.23 + ... + n.2n = (n − 1) 2n+1+2
1.2 + 2.3 + 3.4 + ... + n (n + 1) = \[\frac{n(n + 1)(n + 2)}{3}\]
52n −1 is divisible by 24 for all n ∈ N.
Prove that n3 - 7n + 3 is divisible by 3 for all n \[\in\] N .
Let P(n) be the statement : 2n ≥ 3n. If P(r) is true, show that P(r + 1) is true. Do you conclude that P(n) is true for all n ∈ N?
\[\frac{(2n)!}{2^{2n} (n! )^2} \leq \frac{1}{\sqrt{3n + 1}}\] for all n ∈ N .
\[\text{ Let } P\left( n \right) \text{ be the statement } : 2^n \geq 3n . \text{ If } P\left( r \right) \text{ is true, then show that } P\left( r + 1 \right) \text{ is true . Do you conclude that } P\left( n \right)\text{ is true for all n } \in N?\]
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 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:
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
8 + 17 + 26 + … + (9n – 1) = `"n"/2(9"n" + 7)`
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
`[(3, -4),(1, -1)]^"n" = [(2"n" + 1, -4"n"),("n", -2"n" + 1)], ∀ "n" ∈ "N"`
Answer the following:
Prove by method of induction 152n–1 + 1 is divisible by 16, for all n ∈ N.
Prove by induction that for all natural number n sinα + sin(α + β) + sin(α + 2β)+ ... + sin(α + (n – 1)β) = `(sin (alpha + (n - 1)/2 beta)sin((nbeta)/2))/(sin(beta/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.
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:
n3 – 7n + 3 is divisible by 3, 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:
2 + 4 + 6 + ... + 2n = n2 + n for all natural numbers n.
Prove the statement by using the Principle of Mathematical Induction:
1 + 5 + 9 + ... + (4n – 3) = n(2n – 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 `1/(n + 1) + 1/(n + 2) + ... + 1/(2n) > 13/24`, for all natural numbers n > 1.
