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Sum of n Terms

If Sn=a1+a2+a...

Question

If $S_{n} = a_{1} + a_{2} + a_{3} + \dots + a_{n},$ where $a_{1} = 1$ and $a_{n} = 2 a_{n - 1} + 1,$ for $n \geq 2$ , then the value $S_{50}$ is

A

2^{51} - 52

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B

2^{51} - 50

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C

2^{50} - 50

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D

2^{50} - 52

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Solution

The correct option is A $2^{51} - 52$
$a_{1} = 1 a_{2} = 2 \times 1 + 1 = 3 a_{3} = 2 \times 3 + 1 = 7$
Now,
$S_{n} = a_{1} + a_{2} + a_{3} + \dots + a_{n} = 1 + 3 + 7 + \dots + a_{n} = (2 - 1) + (2^{2} - 1) + (2^{3} - 1) + \dots + (2^{n} - 1) = (2 + 2^{2} + 2^{3} + \dots + 2^{n}) - n = 2 (\frac{2^{n} - 1}{2 - 1}) - n = 2^{n + 1} - n - 2$
$∴ S_{50} = 2^{51} - 52$

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Similar questions

S_{n} = a_{1} + a_{2} + a_{3} + \dots + a_{n},

a_{1} = 1

a_{n} = 2 a_{n - 1} + 1,

n \geq 2

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S_{50}

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