Prove the following by using the first principle
$1 + 3 + 3^{2} + . . . . . . + 3^{n - 1} = \frac{(3^{n} - 1)}{2}$ .

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Solution

$(a) c h e c k p (1) L H S = 1 R H S = (\frac{3^{1} - 1}{2}) = (\frac{2}{2}) = 1 L H S = R H S (b) l e t P (K) b e t r u e, t h e n 1 + 3 + 3^{2} - - - - - - - - - - + 3^{k - 1} = (\frac{3^{k} - 1}{2}) n o w f o r p (k + 1) L H S = 1 + 3 + 3^{2} - - - - - - - + 3^{K - 1} + 3^{k} a G p w i t h (k + 1) t e r m s = (\frac{1 \cdot (3^{k + 1} - 1)}{3 - 1}) = (\frac{3^{k + 1}}{2}) R H S = (\frac{3^{k + 1}}{2}) ∴ L H S = R H S H e n c e p (k + 1) i s t r u e, s o p (n) i s T r u e f o r a l l n \in N u s i n g P M I$

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