Advertisements
Advertisements
Question
12 + 22 + 32 + ... + n2 =\[\frac{n(n + 1)(2n + 1)}{6}\] .
Advertisements
Solution
Let P(n) be the given statement.
Now,
\[P(n) = 1^2 + 2^2 + 3^2 + . . . + n^2 = \frac{n(n + 1)(2n + 1)}{6}\]
\[\text{ Step } 1: \]
\[P(1) = 1^2 = \frac{1(1 + 1)(2 + 1)}{6} = \frac{6}{6} = 1\]
\[\text{ Hence, P(1) is true} . \]
\[\text{ Step } 2: \]
\[\text{ Let P(m) be true .} \]
\[\text{ Then,} \]
\[ 1^2 + 2^2 + . . . + m^2 = \frac{m(m + 1)(2m + 1)}{6}\]
\[\text{ We shall now prove that P(m + 1) is true} . \]
\[i . e . , \]
\[ 1^2 + 2^2 + 3^2 + . . . + (m + 1 )^2 = \frac{(m + 1)(m + 2)(2m + 3)}{6}\]
\[ \text{ Now } , \]
\[P(m) = 1^2 + 2^2 + 3^2 + . . . + m^2 = \frac{m(m + 1)(2m + 1)}{6}\]
\[ \Rightarrow 1^2 + 2^2 + 3^2 + . . . + m^2 + (m + 1 )^2 = \frac{m(m + 1)(2m + 1)}{6} + (m + 1 )^2 \left[ \text{ Adding} (m + 1 )^2 \text{ to both sides} \right]\]
\[ \Rightarrow 1^2 + 2^2 + 3^2 + . . . + (m + 1 )^2 = \frac{m(m + 1)(2m + 1) + 6(m + 1 )^2}{6} = \frac{(m + 1)(2 m^2 + m + 6m + 6)}{6} = \frac{(m + 1)(m + 2)(2m + 3)}{6}\]
\[\text{ Hence, P(m + 1) is true } . \]
\[\text{ By the principle of mathematical induction, the given statement is true for all n } \in N .\]
APPEARS IN
RELATED QUESTIONS
Prove the following by using the principle of mathematical induction for all n ∈ N:
`1 + 3 + 3^2 + ... + 3^(n – 1) =((3^n -1))/2`
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:
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:
\[\frac{1}{2 . 5} + \frac{1}{5 . 8} + \frac{1}{8 . 11} + . . . + \frac{1}{(3n - 1)(3n + 2)} = \frac{n}{6n + 4}\]
1.3 + 3.5 + 5.7 + ... + (2n − 1) (2n + 1) =\[\frac{n(4 n^2 + 6n - 1)}{3}\]
a + ar + ar2 + ... + arn−1 = \[a\left( \frac{r^n - 1}{r - 1} \right), r \neq 1\]
n(n + 1) (n + 5) is a multiple of 3 for all n ∈ N.
11n+2 + 122n+1 is divisible by 133 for all n ∈ N.
7 + 77 + 777 + ... + 777 \[{. . . . . . . . . . .}_{n - \text{ digits } } 7 = \frac{7}{81}( {10}^{n + 1} - 9n - 10)\]
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?
Prove by method of induction, for all n ∈ N:
12 + 32 + 52 + .... + (2n − 1)2 = `"n"/3 (2"n" − 1)(2"n" + 1)`
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:
(24n−1) is divisible by 15
Prove by method of induction, for all n ∈ N:
5 + 52 + 53 + .... + 5n = `5/4(5^"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
`[(3, -4),(1, -1)]^"n" = [(2"n" + 1, -4"n"),("n", -2"n" + 1)], ∀ "n" ∈ "N"`
Answer the following:
Prove by method of induction 52n − 22n is divisible by 3, for all n ∈ N
Prove statement by using the Principle of Mathematical Induction for all n ∈ N, that:
22n – 1 is divisible by 3.
Prove by the Principle of Mathematical Induction that 1 × 1! + 2 × 2! + 3 × 3! + ... + n × n! = (n + 1)! – 1 for all natural numbers n.
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`
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:
n(n2 + 5) is divisible by 6, for each natural number n.
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:
2n < (n + 2)! for all natural number 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.
Prove the statement by using the Principle of Mathematical Induction:
1 + 2 + 22 + ... + 2n = 2n+1 – 1 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.
Prove that, sinθ + sin2θ + sin3θ + ... + sinnθ = `((sin ntheta)/2 sin ((n + 1))/2 theta)/(sin theta/2)`, for all n ∈ N.
Show that `n^5/5 + n^3/3 + (7n)/15` is a natural number for all n ∈ N.
Prove that number of subsets of a set containing n distinct elements is 2n, for all n ∈ N.
If xn – 1 is divisible by x – k, then the least positive integral value of k is ______.
By using principle of mathematical induction for every natural number, (ab)n = ______.
