The correct option is B 3^3000 nbsp; 1/2 The given series is a GP with first term nbsp;a=1 and the common ratio r nbsp;=t_2/t_1 nbsp;= 3/1 nbsp;= 3. ...

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Byju's Answer

Standard X

Mathematics

Sum of n Terms

1+31+32+33………...

Question

$1 + 3^{1} + 3^{2} + 3^{3} . . . . . . . . . . + 3^{2999}$ =

$\frac{5^{2999} - 1}{2}$

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$\frac{3^{3000} - 1}{2}$

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$\frac{3^{2999} - 1}{2}$

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$\frac{4^{1999} - 1}{2}$

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Solution

The correct option is B
$\frac{3^{3000} - 1}{2}$

The given series is a GP with first term $a = 1$ and the common ratio $r = \frac{t_{2}}{t_{1}} = \frac{3}{1} = 3.$

The sum to $n$ terms of the GP is given by
$S_{n} = \frac{a (r^{n} - 1)}{r - 1}, where r > 1.$

$∴ 1 + 3^{1} + 3^{2} + 3^{3} . . . . . . . . . . + 3^{2999} = \frac{1 (3^{3000} - 1)}{3 - 1}$

$= \frac{(3^{3000} - 1)}{2}$

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1+31+32+33..........+32999 =

The correct option is B 33000−12 The given series is a GP with first term a=1 and the common ratio r=t2t1=31=3. The sum to n terms of the GP is given by Sn=a(rn−1)r−1 , where r>1. ∴1+31+32+33..........+32999=1(33000−1)3−1 =(33000−1)2

$1 + 3^{1} + 3^{2} + 3^{3} . . . . . . . . . . + 3^{2999}$ =