Question

প্রমাণ কর যে: $\dfrac{4^{n}-1}{2^{n}-1}=2^{n}+1$

Answer 1

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]

Comments

Rules Applied:

• $\left(a^{m}\right)^{n}=a^{mn}$
• $a^2-b^2=(a+b)(a-b)$