Advertisements
Advertisements
Question
7 + 77 + 777 + ... + 777 \[{. . . . . . . . . . .}_{n - \text{ digits } } 7 = \frac{7}{81}( {10}^{n + 1} - 9n - 10)\]
Advertisements
Solution
Let P(n) be the given statement.
Now,
\[P(n): 7 + 77 + 777 + . . . + 777 . . ._{\text{ n digits } } . . . 7 = \frac{7}{81}( {10}^{n + 1} - 9n - 10)\]
\[\text{ Step(1): } \]
\[P(1) = 7 = \frac{7}{81}( {10}^2 - 9 - 10) = \frac{7}{81} \times 81 \]
\[\text{ Thus, P(1) is true } . \]
\[\text{ Step } 2: \]
\[\text{ Let P(m) be true } . \]
\[\text{ Then,} \]
\[7 + 77 + 777 + . . . + 777 . . ._{\text{ m digits} } . . . 7 = \frac{7}{81}( {10}^{m + 1} - 9m - 10)\]
\[\text{ We need to show that P(m + 1) is true whenever P(m) is true} . \]
Now, P(m + 1) = 7 + 77 + 777 +....+ 777...(m + 1) digits...7
\[\text{ This is a geometric progression with } n = m + 1 . \]
\[ \therefore \text{ Sum } P(m + 1): \]
\[ = \frac{7}{9}\left[ 9 + 99 + 999 + . . . \left( m + 1 \right)term \right]\]
\[ = \frac{7}{9}\left[ \left( 10 - 1 \right) + \left( 100 - 1 \right) + . . . (m + 1) \text{ term } \right]\]
\[ = \frac{7}{9}\left[ 10 + 100 + 1000 + . . . (m + 1) \text { term } - (1 + 1 + 1 . . . m + 1\text{ times} . . . + 1 \right]\]
\[ = \frac{7}{9}\left[ \frac{10\left( {10}^{m + 1} - 1 \right)}{9} - m + 1 \right]\]
\[ = \frac{7}{81}\left[ {10}^{m + 2} - 9m - 19 \right]\]
\[\text{ Thus, P(m + 1) is true } . \]
\[\text{ By the principle of mathematical induction, } P\left( n \right)\text{ is true for all n } \in N . \]
APPEARS IN
RELATED QUESTIONS
Prove the following by using the principle of mathematical induction for all n ∈ N: 1.2.3 + 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:
`(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: n (n + 1) (n + 5) is a multiple of 3.
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.
Prove the following by using the principle of mathematical induction for all n ∈ N: 41n – 14n is a multiple of 27.
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 "n2 + n is even", and if P (r) is true, then P (r + 1) is 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.
\[\frac{1}{1 . 2} + \frac{1}{2 . 3} + \frac{1}{3 . 4} + . . . + \frac{1}{n(n + 1)} = \frac{n}{n + 1}\]
\[\frac{1}{1 . 4} + \frac{1}{4 . 7} + \frac{1}{7 . 10} + . . . + \frac{1}{(3n - 2)(3n + 1)} = \frac{n}{3n + 1}\]
1.2 + 2.3 + 3.4 + ... + n (n + 1) = \[\frac{n(n + 1)(n + 2)}{3}\]
32n+7 is divisible by 8 for all n ∈ N.
32n+2 −8n − 9 is divisible by 8 for all n ∈ N.
72n + 23n−3. 3n−1 is divisible by 25 for all n ∈ 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:
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)`
Prove by method of induction, for all n ∈ N:
(23n − 1) is divisible by 7
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:
`[(1, 2),(0, 1)]^"n" = [(1, 2"n"),(0, 1)]` ∀ n ∈ N
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 152n–1 + 1 is divisible by 16, for all n ∈ N.
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:
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:
For any natural number n, xn – yn is divisible by x – y, where x and y are any integers with x ≠ y.
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:
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 that, cosθ cos2θ cos22θ ... cos2n–1θ = `(sin 2^n theta)/(2^n sin theta)`, 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.
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.
