Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
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1 + 3 + 32 + ... + 3n−1 = 3 N − 1 2 - Mathematics

1 + 3 + 32 + ... + 3n−1 = \[\frac{3^n - 1}{2}\]

 
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Solution

Let P(n) be the given statement.
Now,

\[P(n) = 1 + 3 + 3^2 + . . . + 3^{n - 1} = \frac{3^n - 1}{2}\]

\[\text{ Step 1} : \]

\[P(1) = 1 = \frac{3^1 - 1}{2} = \frac{2}{2} = 1\]

\[\text{ Hence, P(1) is true } . \]

\[\text{ Step 2:}  \]

\[\text{ Let P(m) is true .}  \]

\[\text{ Then } , \]

\[1 + 3 + 3^2 + . . . + 3^{m - 1} = \frac{3^m - 1}{2}\]

\[\text{ We shall prove that P(m + 1) is true }  . \]

\[\text{ That is, }  \]

\[1 + 3 + 3^2 + . . . + 3^m = \frac{3^{m + 1} - 1}{2}\]

\[\text{ Now, we have: }  \]

\[1 + 3 + 3^2 + . . . + 3^{m - 1} = \frac{3^m - 1}{2}\]

\[ \Rightarrow 1 + 3 + 3^2 + . . . + 3^{m - 1} + 3^m = \frac{3^m - 1}{2} + 3^m \left[ \text{ Adding } 3^m \text{ to both sides } \right]\]

\[ \Rightarrow 1 + 3 + 3^2 + . . . + 3^m = \frac{3^m - 1 + 2 \times 3^m}{2} = \frac{3^m (1 + 2) - 1}{2} = \frac{3^{m + 1} - 1}{2}\]

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

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

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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 12 Mathematical Induction
Exercise 12.2 | Q 3 | Page 27
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