7 + 77 + 777 + ... + 777 \[{. . . . . . . . . . .}_{n - \text{ digits } } 7 = \frac{7}{81}( {10}^{n + 1} - 9n - 10)\]

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 \[\text{ This is a geometric progression with } n = m + 1 . \] \[ \therefore \text{ Sum } P(m + 1): \] \[ = \frac{7}{9}\left[ 9 + 99 + 999 + . . . \left( m + 1 \right)term \right]\] \[ = \frac{7}{9}\left[ \left( 10 - 1 \right) + \left( 100 - 1 \right) + . . . (m + 1) \text{ term } \right]\] \[ = \frac{7}{9}\left[ 10 + 100 + 1000 + . . . (m + 1) \text { term } - (1 + 1 + 1 . . . m + 1\text{ times} . . . + 1 \right]\] \[ = \frac{7}{9}\left[ \frac{10\left( {10}^{m + 1} - 1 \right)}{9} - m + 1 \right]\] \[ = \frac{7}{81}\left[ {10}^{m + 2} - 9m - 19 \right]\] \[\text{ Thus, P(m + 1) is true } . \] \[\text{ By the principle of mathematical induction, } P\left( n \right)\text{ is true for all n } \in N . \]

7 + 77 + 777 + ... + 777 . . . . . . . . . . . N − Digits 7 = 7 81 ( 10 N + 1 − 9 N − 10 ) - Mathematics

Question

7 + 77 + 777 + ... + 777 ${. . . . . . . . . . .}_{n - digits} 7 = \frac{7}{81} (10^{n + 1} - 9 n - 10)$

Solution

Let P(n) be the given statement.
Now,

$P (n) : 7 + 77 + 777 + . . . + 777 . . ._{n digits} . . . 7 = \frac{7}{81} (10^{n + 1} - 9 n - 10)$
$Step(1):$
$P (1) = 7 = \frac{7}{81} (10^{2} - 9 - 10) = \frac{7}{81} \times 81$
$Thus, P(1) is true .$
$Step 2 :$
$Let P(m) be true .$
$Then,$
$7 + 77 + 777 + . . . + 777 . . ._{m digits} . . . 7 = \frac{7}{81} (10^{m + 1} - 9 m - 10)$
$We need to show that P(m + 1) is true whenever P(m) is true .$

Now, P(m + 1) = 7 + 77 + 777 +....+ 777...(m + 1) digits...7

$This is a geometric progression with n = m + 1 .$

$∴ Sum P (m + 1) :$

$= \frac{7}{9} [9 + 99 + 999 + . . . (m + 1) t e r m]$

$= \frac{7}{9} [(10 - 1) + (100 - 1) + . . . (m + 1) term]$

$= \frac{7}{9} [10 + 100 + 1000 + . . . (m + 1) term - (1 + 1 + 1 . . . m + 1 times . . . + 1]$

$= \frac{7}{9} [\frac{10 (10^{m + 1} - 1)}{9} - m + 1]$

$= \frac{7}{81} [10^{m + 2} - 9 m - 19]$

$Thus, P(m + 1) is true .$

$By the principle of mathematical induction, P (n) is true for all n \in N .$

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Principle of Mathematical Induction

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Q 31 Q 30 Q 32

Chapter 12: Mathematical Induction - Exercise 12.2 [Page 28]

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RD Sharma Mathematics [English] Class 11

Chapter 12 Mathematical Induction
Exercise 12.2 | Q 31 | Page 28

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