Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
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7 + 77 + 777 + ... + 777 . . . . . . . . . . . N − Digits 7 = 7 81 ( 10 N + 1 − 9 N − 10 ) - Mathematics

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

 
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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 

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

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

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

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

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

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

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

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

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

 

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APPEARS IN

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