GMAT CLUB OLYMPICS: If a + a/(1 + 2)

Question

If  a + \frac{a}{(1 + 2)} + \frac{a}{(1 + 2 + 3)} + ... + \frac{a}{(1 + 2 + 3 + ... + 101)} = 101 , then what is the value of a ?

A. 50
B. 51
C. 52
D. 55
E. 100

M37-10

Bunuel · Accepted Answer

Official Solution:

If  a + \frac{a}{1 + 2} + \frac{a}{1 + 2 + 3} + ... + \frac{a}{1 + 2 + 3 + ... + 101} = 101 , then what is the value of  a  ?

A.  50 
B.  51 
C.  52 
D.  55 
E.  100

a + \frac{a}{1 + 2} + \frac{a}{1 + 2 + 3} + ... + \frac{a}{1 + 2 + 3 + ... + 101} = 101 ;

Factor out  a :  a(1 + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + ... + \frac{1}{1 + 2 + 3 + ... + 101}) = 101 ;

In the denominators we have the sum of the first  n  positive integers, which equals to  \frac{n(n+1)}{2} .
 
Re-write the denominators according to the above:  a(1 + \frac{1}{\frac{2(2+1)}{2}} + \frac{1}{\frac{3(3+1)}{2}} + ... + \frac{1}{\frac{101(101+1)}{2}}) = 101 ;

Simplify:  a(1 + \frac{2}{2(2 +1 )} + \frac{2}{3(3+1)} + ... + \frac{2}{101(101+1)}) = 101 ;

Factor out 2:  2a(\frac{1}{2} + \frac{1}{2(2 +1 )} + \frac{1}{3(3+1)} + ... + \frac{1}{101(101+1)}) = 101 ;

2a(\frac{1}{2} + \frac{1}{2*3} + \frac{1}{3*4} + ... + \frac{1}{101*102}) = 101 ;

2a((1-\frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + ... + (\frac{1}{101} - \frac{1}{102})) = 101 ;
 
Notice that everything in the brackets except the first and last number will cancel out and we'll get:  2a(1 - \frac{1}{102}) = 101 ;

2a(\frac{101}{102}) = 101 ;

a=51 .

Answer: B

Berdiyor · Answer

Hello everyone

Here is my solution. 
The correct answer is 51 which is B

Kinshook · Answer

Asked: If  a + \frac{a}{(1 + 2)} + \frac{a}{(1 + 2 + 3)} + ... + \frac{a}{(1 + 2 + 3 + ... + 101)} = 101 , then what is the value of a ?

t_n = \frac{a}{(1+2+....n)} = \frac{2a}{n(n+1)} = \frac{2a{(n+1) -n}}{n(n+1)} = 2a (\frac{1}{n} - \frac{1}{n+1})

a + \frac{a}{(1 + 2)} + \frac{a}{(1 + 2 + 3)} + ... + \frac{a}{(1 + 2 + 3 + ... + 101)} = 101 
 2a(1 -1/2) + 2a(1/2 - 1/3) + 2a(1/3 - 1/4) + ......+ 2a(1/101 - 1/102) = 101 
 2a(1 - 1/102) = 101 
 2a(101/102) = 101 
 a = 51

IMO B

Vjajo · Answer

Here the Nth term is = a/(1+2+3+..+n)
 
Sum of first n natural nos. = n(n+1)/2
Therefore the nth term becomes = 2a/(n(n+1))
 =2a 
On summation of these terms till 101, all the middle terms get cancelled and we are left with 2a 
Now, 2a  = 101
=> 2a*101/102 = 101
=> a = 51

Hence Option B is the correct answer.

hnguye10 · Answer

We have:
1+2+...+n = (n+1)n/2 
=> 1/(1+2+...+n) = 2/(n+1)n = 2*(1/n - 1/(n+1))

Thus we have:
a/1 + a/(1+2) + ... + a/(1+2+...+101) = 101
=> a*(1/1 + 1/(1+2) + ... + 1/(1+2+...+101)) = 101
=> 2a*(1/1 - 1/2 + 1/2 - 1/3 +... + 1/101 - 1/102) = 101
=> 2a*(1 - 1/102) = 101
=>2a*101/102 = 101
=> 2a = 102
=> a = 51

The answer is B

huongguyen · Answer

a(1 + 1/3 +1/6 +1/10 + ...1/(1+...101) = 101 because the sum (1/3 +1/6 +1/10 + ...1/(1+...10)) < 1 => 1 + sum < 2 => a(1 + sum) < 2a <=> 101 < 2a <=> 101/2 < a <=> 50.5 < a => a = 51

chetan2u · Answer

The moment you see the formula of nth number, it should tell you we are looking at a sequence ..
Also, the denominator seems to be increasing in a certain pattern. So, let us try to find out.

Here the nth term is  \frac{a}{\frac{n(n+1)}{2}}=\frac{2a}{n(n+1)}=2a(\frac{1}{n}-\frac{1}{n+1})

a + \frac{a}{(1 + 2)} + \frac{a}{(1 + 2 + 3)} + ... + \frac{a}{(1 + 2 + 3 + ... + 101)} = 101

2a(\frac{1}{1*2}+\frac{1}{(1 + 2)*2} + \frac{1}{(1 + 2 + 3)*2} + ... + \frac{1}{(1 + 2 + 3 + ... + 101)*2} = 101

\frac{1}{(1 )*2} =\frac{1}{1*2}= \frac{1}{1}-\frac{1}{2} ..

\frac{1}{(1 + 2)*2} =\frac{1}{3*2}= \frac{1}{2}-\frac{1}{3} .

\frac{1}{(1 + 2+3)*2} = \frac{1}{3}-\frac{1}{4} ..
.....
  \frac{1}{(1 +2+..+101)*2} =\frac{1}{101*102}= \frac{1}{101}- \frac{1}{102} .

ADD all 
 2a(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}..... \frac{1}{100}-\frac{1}{101}+\frac{1}{101}-\frac{1}{102}) = 2a(1-\frac{1}{102}) =2a( \frac{101}{102})=101 
 a=51

B

heisenberg26 · Answer

You can solve it by using the knowledge of just this one  formula  -  Sum of n consecutive terms from 1 to n is n(n+1)/2

Step 1  - Take the common term a out of the expression : a * (1 +  \frac{1}{1+2}   +  \frac{1}{1+2+3}   + ... +  \frac{1}{1+2+3+..101}   ) = 101

Step 2  - Substitute the terms with the formula : a x (1 +  \frac{2}{2 * 3}   +  \frac{2}{3 * 4}   + ... +  \frac{2}{101 * 102}   ) = 101

Step 3  - Write 1 as 2/2 to create standard form : a * ( \frac{2}{1 * 2}   +  \frac{2}{2 * 3}   +  \frac{2}{3 * 4}   + ... +  \frac{2}{101 * 102}   ) = 101

Step 4  - Take common term 2 out : a * 2 * ( \frac{1}{1 * 2}   +  \frac{1}{2 * 3}   +  \frac{1}{3 * 4}   + ... +  \frac{1}{101 * 102}   ) = 101

Now,  \frac{1}{n * (n+1)}   can be written as  \frac{(n+1) - n }{n * (n+1)}   which can further be broken down into (  \frac{1}{n}   -  \frac{1}{(n+1)}   )

Step 5  - Substitute the expression with the from above : a * 2 * ( \frac{1}{1}   -  \frac{1}{2}   +  \frac{1}{2}   -  \frac{1}{3}   +  \frac{1}{3}   + ... -  \frac{1}{101}   +  \frac{1}{101}   -  \frac{1}{102}  ) = 101

We can observe that all terms in the fractional expression cancel out except the first and the last terms.

Step 6  - Cancel out terms in the fractional expression : a * 2 * ( \frac{1}{1}   -  \frac{1}{102}  ) = 101

Step 7  - Find out a by cancelling out common term 101 on LHS and RHS : a * 2 * ( \frac{101}{102}   ) = 101

( a *2 *  \frac{1}{102}   )= 1

Answer  ->  a=51

Rebaz · Answer

Bunuel

May i ask you how you go from 1/101(101+1) to 1/101∗104?

Why do you change the (101+1),which equal to 102 to 104?????

Thanks in advance!

Bunuel · Answer

It was a typo. Edited. Thank you!

Diegoram17 · Answer

I just wonder if its possible to answer this kind of questions in less than 2 minutes?

GMAT CLUB OLYMPICS: If a + a/(1 + 2)

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