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

# 1 + 3 + 32 + ... + 3n−1 = 3 N − 1 2 - Mathematics

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

#### 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