sritamasia
If \(R\) = \( \frac{{30^{65} - 29^{65}}}{{30^{64} + 29^{64}}}\), then R?
A. 0 < R < 0.1
B. 0.1 < R < 0.5
C. 0.5 < R < 1.0
D. R > 1.0
E. R < - 1.0
Without getting into any algebraic identities or formulas, a quick way keeping the options in mind would be :-
Get the numerator in similar terms as the denominator.
\( \frac{{30^{65} - 29^{65}}}{{30^{64} + 29^{64}}}\)
\( \frac{30*{30^{64} -29* 29^{64}}}{{30^{64} + 29^{64}}}\)
We have 30 times in \(30^{64}\), so let us cancel out and add 29 times \(30^{64}\). Similarly for other term.
\( \frac{(30*30^{64}-29*30^{64}+29*30^{64})+( -29* 29^{64}+30*29^{64}-30*29^{64})}{{30^{64} + 29^{64}}}\)
\( \frac{(30^{64}+29*30^{64}) +(29^{64}-30*29^{64})}{{30^{64} + 29^{64}}}\)
\( \frac{(30^{64}+ 29^{64} )+(29*30^{64}-30*29^{64})}{{30^{64} + 29^{64}}}\)
\(\frac{ {30^{64} + 29^{64}}}{{30^{64} + 29^{64}}}+ \frac{29*30*(30^{63}) -30*29*(29^{63})}{{30^{64} + 29^{64}}}\)
= 1+something positive
= >1
You could also find pattern in \(\frac{30^a-29^a}{30^{a-1}+29^{a-1}}\)
a=1 gives value of expression as 1/2
a=2 gives value as 1
a=3 gives value >1
So we can see that value increases as a increases, and we have to find value when a=65.
Thus, surely >1.
D