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

Given that,

• $u_{n+1} = 2 u_{n} + 1 (n=0,1,2,\dots)$
• $u_{n} = 0$

Now, put the various values of $n$ and see the pattern.

• $n=0 \Rightarrow u_{1} = 2u_{0} + 1 = 1= 2^{1}-1$
• $n=1 \Rightarrow u_{2} = 2u_{1} + 1 = 3= 2^{2}-1$
• $n=2 \Rightarrow u_{3} = 2u_{2} + 1 = 7= 2^{3}-1$
• $n=3 \Rightarrow u_{4} = 2u_{3} + 1 = 15= 2^{4}-1$

Now, we can generalize the pattern.

$u_{n} = 2^{n} – 1$

$\Rightarrow u_{10} = 2^{10} – 1$

$\Rightarrow u_{10} = 1024 – 1$

$\Rightarrow \boxed{u_{10} = 1023}$

Correct Answer $:1023$

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