# CAT 2014 | Question: 45

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Let $a_{n+1}= 2 a_{n}+1 ({n}=0, 1, 2,\dots)$ and $a_{0}=0$. Then $a_{10}$ nearest to.

1. $1023$
2. $2047$
3. $4095$
4. $8195$

edited

Given that, $a_{n+1} = 2a_{n}+1; n=0,1,2,3, \dots , a_{0} = 0$

put the various values of $n$ in equation $(1)$ and observe the pattern.

• $n=0: a_{1} = 2a_{0}+1 = 2(0)+1=1 = 2^{1}-1$
• $n=1: a_{2} = 2a_{1}+1 = 2(1)+1=3 = 2^{2}-1$
• $n=2: a_{3} = 2a_{2}+1 = 2(3)+1=7 = 2^{3}-1$
• $n=3: a_{4} = 2a_{3}+1 = 2(7)+1=15 = 2^{4}-1$
• $\vdots$
• $n=k : a_{k+1} = 2a_{k}+1 = 2^{k+1}-1$
• $n=9 : a_{10} = 2a_{9}+1 = 2^{10}-1 = 1024-1 = 1023$
• Correct Answer $: \text{A}$
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