The value of \(\frac{5.(25)^{n+1}+25.(5)^{2n+1}}{25.(5)^{2n}-105(25)^

1.	The value of \(\frac{5.(25)^{n+1}+25.(5)^{2n+1}}{25.(5)^{2n}-105(25)^{n-1}} \) is(a) 0 (b) 1 (c) \(6\frac14\)(d) \(5\frac14\)
Answer» (c) \(6\frac14\) Given exp. = \(\frac{5.(5^2)^{n+1}+5^2.5^{2n+1}}{5^2.5^{2n}-21\times5(5^2)^{n-1}} \) = \(\frac{5^{2n+3}+5^{2n+1}}{5^{2n+2}-21\times5^{2n-1}}\) = \(\frac{5^{2n+1}(5^2+1)}{5^{2n-1}(5^3-21)}\) = \(\frac{5^{(2n+1)-(2n-1)}\times26}{(125-21)}\) = \(\frac{5^2\times26}{104}=\frac{25}{4}=6\frac14.\)

The value of \(\frac{5.(25)^{n+1}+25.(5)^{2n+1}}{25.(5)^{2n}-105(25)^{n-1}} \) is(a) 0 (b) 1 (c) \(6\frac14\)(d) \(5\frac14\)

Answer»

(c) \(6\frac14\)

Given exp. = \(\frac{5.(5^2)^{n+1}+5^2.5^{2n+1}}{5^2.5^{2n}-21\times5(5^2)^{n-1}} \)

= \(\frac{5^{2n+3}+5^{2n+1}}{5^{2n+2}-21\times5^{2n-1}}\) = \(\frac{5^{2n+1}(5^2+1)}{5^{2n-1}(5^3-21)}\)

= \(\frac{5^{(2n+1)-(2n-1)}\times26}{(125-21)}\) = \(\frac{5^2\times26}{104}=\frac{25}{4}=6\frac14.\)