প্রমাণ কর যে: $\dfrac{4^{n}-1}{2^{n}-1}=2^{n}+1$
প্রমাণ কর যে: $\dfrac{4^{n}-1}{2^{n}-1}=2^{n}+1$
1 টি উত্তর পাওয়া গেছে
Left Hand Side,
$=\dfrac{4^{n}-1}{2^{n}-1}$
$=\dfrac{\left(2^2\right)^{n}-1}{2^{n}-1}$
$=\dfrac{2^{2n}-1}{2^{n}-1}$
$=\dfrac{\left(2^{n}\right)^2-1^2}{2^{n}-1}$
$=\dfrac{\left(2^{n}+1\right)\left(2^{n}-1\right)}{\left(2^{n}-1\right)}$
$=\left(2^{n}+1\right)$
$=$ Right Hand Side [Proved]
$=\dfrac{4^{n}-1}{2^{n}-1}$
$=\dfrac{\left(2^2\right)^{n}-1}{2^{n}-1}$
$=\dfrac{2^{2n}-1}{2^{n}-1}$
$=\dfrac{\left(2^{n}\right)^2-1^2}{2^{n}-1}$
$=\dfrac{\left(2^{n}+1\right)\left(2^{n}-1\right)}{\left(2^{n}-1\right)}$
$=\left(2^{n}+1\right)$
$=$ Right Hand Side [Proved]
Rules Applied: