Show that 1 - 3 - 32 - - 3n - 1 - 3n - 1-2

1+3+3^2+.....+3^n 1 This is a GP series #160; a=1,r=3 sum #160; of GP series #160; =a(r^n 1)r 1 =(3^n 1)3 1 ...

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Standard X

Mathematics

Properties of GP

Show that 1...

Question

Show that $1 + 3 + 3^{2} + . . . . . 3^{n - 1} = \frac{3^{n} - 1}{2}$

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Solution

$1 + 3 + 3^{2} + . . . . . + 3^{n - 1}$
This is a GP series
$a = 1, r = 3$
sum of GP series $= \frac{a (r^{n} - 1)}{r - 1}$
$= \frac{(3^{n} - 1)}{3 - 1}$

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Q. Prove that :

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$1 + 3 + 3^{2} + \dots + 3^{n - 1} = \frac{(3^{n} - 1)}{2}$

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Show that 1+3+32+.....3n−1=3n−12

1+3+32+.....+3n−1This is a GP series a=1,r=3sum of GP series =a(rn−1)r−1=(3n−1)3−1

Show that $1 + 3 + 3^{2} + . . . . . 3^{n - 1} = \frac{3^{n} - 1}{2}$

$1 + 3 + 3^{2} + . . . . . + 3^{n - 1}$
This is a GP series
$a = 1, r = 3$
sum of GP series $= \frac{a (r^{n} - 1)}{r - 1}$
$= \frac{(3^{n} - 1)}{3 - 1}$