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# If $\;a_{1}=1\;,a_{n+1}=2a_{n}+1$ then , $\;a_{n+1}$ equals

$(a)\;2^n-2\qquad(b)\;2^{n+1}-2\qquad(c)\;2^n+1\qquad(d)\;2^{n+1}-1$

Can you answer this question?

Answer : (d) $\;2^{n+1}-1$
Explanation : $\;a_{n+1}-a_{n}=2(a_{n}-a_{n-1})$
$=2^2\;(a_{n-1}-a_{n-2})$
$=2^3\;(a_{n-2}-a_{n-3})$
$=2^{n}$
$a_{n+1}=2^n+a_{n}$
$=2^{n}+2^{n-1}+a_{n-1}$
$=2^n+2^{n-1}+2^{n-2}+a_{n-2}....$
$=2^n+2^{n-1}+....\;2+a_{1}$
$=2^{n+1}-1\;.$
answered Jan 22, 2014 by