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Question Stats:
35% (02:10) correct
65%
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based on 96
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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
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
I got this wrong the 1st time, but then worked out based on OA
(a^n-b^n) = (a-b)*(a^n-1+....b^n-1). Meaning the 2nd paranthesis is always +ve and greater than a^n-1 + b^n-1.
30^65-29^65 = (30-29)*(30^64+....29^64). The 2nd part is clearly greater than (30^64+29^64)
Hence, R should be greater than 1. Option D
Posted from my mobile device
(a^n-b^n) = (a-b)*(a^n-1+....b^n-1). Meaning the 2nd paranthesis is always +ve and greater than a^n-1 + b^n-1.
30^65-29^65 = (30-29)*(30^64+....29^64). The 2nd part is clearly greater than (30^64+29^64)
Hence, R should be greater than 1. Option D
Posted from my mobile device
SushantSaini
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the27s
I think this generalized formula is not correct.it doesn't even hold good for x3-y3
Abhishek009
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sritamasia
\( \frac{{30^{65} - 29^{65}}}{{30^{64} + 29^{64}}}\)
Or, \( \frac{(30^{64} + 29^{64})(30 - 29)}{(30^{64} + 29^{64})}\)
Or, \(30 - 29 = 1\), Answer must be (D)
_________________
(30-29)*(30^64+29^64)=
30^65-29^65-29*30^64+30*29^64
So
(30^65-29^65)= (30-29)*(30^64+29^64)+29*30^64-30*29^64
Dividing the above by (30^64+29^64) =
1+ (29*30^64-30*29^64)/(30^64+29^64)
Which equals
1 + (30*29)*(30^63-29^63)/(30^64+29^64)
Which is 1 plus a positive number, so R>1
Posted from my mobile device
30^65-29^65-29*30^64+30*29^64
So
(30^65-29^65)= (30-29)*(30^64+29^64)+29*30^64-30*29^64
Dividing the above by (30^64+29^64) =
1+ (29*30^64-30*29^64)/(30^64+29^64)
Which equals
1 + (30*29)*(30^63-29^63)/(30^64+29^64)
Which is 1 plus a positive number, so R>1
Posted from my mobile device
sritamasia
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
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