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Please help on Problem Solving

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Please help on Problem Solving

by teenmark » Mon Dec 21, 2009 4:52 pm
1. For every integer K from 1 to 10, inclusive, the Kth term of a certain sequence is given by (-1)^k+1 . (1/2^k). If T is the sum of the first 10 terms in the sequence, then T is
a. greater than 2
b. between 1 and 2
c. between 1/2 and 1
d. between 1/4 and 1/2
e. less than 1/4

2. For which of the following function is f(a+b)=f(a)+f(b) for all positive numbers a and b?
a.f(x)= x^2
b.f(x)= x+1
c.f(x)= -3x
d.f(x)= square root of x
e.f(x)=2/x

I can not solve these two problems. Please help me, thank you!!!
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by papgust » Mon Dec 21, 2009 8:54 pm
For 1st, check this out
https://www.beatthegmat.com/gmat-prep-qu ... tml#206219

For 2nd,
let's plugin values for each choice and see which choice satisfies f(a+b)=f(a)+f(b).

A. Assume f(a)=a^2, f(b)=b^2. f(a+b)=(a+b)^2 (RHS), which is not equal to a^2 + b^2 [f(a) + f(b)]
B. Assume f(a)=a+1, f(b)=b+1. f(a+b)=a+b+1 (RHS), which is not equal to (a+1)+(b+1) = a+b+2 [f(a) + f(b)]
C. Assume f(a)= -3a, f(b)= -3b. f(a+b)= -3(a+b) = (-3a) + (-3b) which is equal to f(a) + f(b). This choice satisfies.

By this time, you can just end this question by ticking C. But i'm just showing you other 2 choices as well.

D. Assume f(a)= sqrt(a), f(b)=sqrt(b). f(a+b) = sqrt(a+b) (RHS), which is not equal to sqrt(a)+sqrt(b) [f(a) + f(b)]
E. Assume f(a)=2/3, f(b) = 2/3. f(a+b)=2/3 (RHS), which is not equal to 2/3 + 2/3 [f(a) + f(b)]

So, its C.
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by teenmark » Mon Dec 21, 2009 10:17 pm
thank you so much...dang, you re very smart!!!
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by Mylogin » Fri Dec 25, 2009 10:15 pm
K=1 => 1/2
K=2 => -1/4
K=3 => 1/8
K=4 => -1/16

One way is to use GP and solve it, other one which I found time saving was

1/2 - 1/4 + 1/8 - 1/16 ...

first 2 terms 1/2 - 1/4 = 1/4
Rest terms 1/8 - 1/16 .... < 1/4

Hence answer is between 1/4 & 1/2, i.e. D
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