Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

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

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

#### 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

$\text{ This is a geometric progression with } n = m + 1 .$

$\therefore \text{ Sum } P(m + 1):$

$= \frac{7}{9}\left[ 9 + 99 + 999 + . . . \left( m + 1 \right)term \right]$

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

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

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

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

$\text{ Thus, P(m + 1) is true } .$

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

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 12 Mathematical Induction
Exercise 12.2 | Q 31 | Page 28