For positive integers 'n', it is known that 1 + 3 + 5 + ... + (2*n - 1

Question

For positive integers 'n', it is known that 1 + 3 + 5 + ... + (2*n - 1) =  n^2  .

Evaluate  \frac{1+3+5+ ... + 739}{741+743+ ... + 1479}

A. 1/2
B. 1/3
C. 1/4
D. 1/5 
E. 1/6

houston1980 · Accepted Answer

I use Pattern Recognition.

(1+3)/(5+7) = 4/12 = 1/3

OR (1+3+5)/(7+9+11) = 9/27= 1/3

Therefore  Option B  is right.

chetan2u · Answer

SUM and AVERAGE

\frac{1+3+5+ ... + 739}{741+743+ ... + 1479}

There are 370 terms in both the denominator and numerator, so we are to divide the averages of both.

\frac{1+3+5+ ... + 739}{741+743+ ... + 1479}=\frac{\frac{1+739}{2}*370}{\frac{741+1479}{2}*370}=\frac{740}{2220}=\frac{1}{3}

FORMULA

\frac{1+3+5+ ... + 739}{741+743+ ... + 1479}=\frac{1+3+5+ ... + 739}{1+2+.....737+739+741+743+ ... + 1479-(1+3+5...+737+739)}

=\frac{1+3+5+ ... + (2*370-1)}{1+2+ ... + (2*740-1)-(1+2+....2*370-1)}=\frac{370^2}{740^2-370^2}

=\frac{370^2}{(370*2)^2-370^2}=\frac{370^2}{370^2(2^2-1)}=\frac{1}{3}

B

ahamd · Answer

Numerator :

1+3+5+....+739

2n-1=739
n=370

The numerator can be written as 370^2

Denominator:

741+743+....+1479
=(1+3+5+.....1479)-(1+3+5+...739)

For the first bracketed portion,

2n-1=1479
n=740

Hence the denominator can be written as 740^2-370^2

Now,
370^2/(740^2-370^2)

=370^2/(740+370)(740-370)
=370^2/(1110)(370)
=370/1110
=1/3
Option B

gmatophobia · Answer

\frac{1+3+5+ ... + 739}{741+743+ ... + 1479}

\frac{1+3+5+ ... + 739}{(740+1) + (740 + 3) + ... + (740 + 739)}

Let's solve  1+3+5+ ... + 739

Number of terms =  \frac{739 - 1 }{ 2}  + 1 = 370

Therefore as per the given equation,  1+3+5+ ... + 739 = 370^2

\frac{1+3+5+ ... + 739}{ (740 + 740 + 740 ... 	ext{370 times}) + (1 + 3 + 5 + ...739) }

\frac{370^2}{ 740 * 370 + 370^2 }

\frac{370^2}{ 370(370 + 740) }

\frac{370}{ 1110 }

\frac{1}{3}

Option B

Tukkebaaz · Answer

Okay so the problem can be solved in <1min 1 + 3 + 5 + ... + (2*n - 1) = n^2 -----Method 1 ] Method 2 below: Numerator = 1 + 3 + 5 + ... + (2*n - 1) = 1+3+5+...+739; where n(num) = 740/2 Denominator = (1+3+...1479) - Numerator ; Here n(denom) = 1480/2 Basically what we have now is (740/2)^2 / Lets take (740/2)^2 common and cancel from num and denom We have 1/(2^2-1) = 1/3. Either methods are good to solve, I went with method 02 as it gives me confidence that I have arrived at my answer without a shadow of doubt. I believe that it is always good to come up with multiple methods of solving during practice to best prepared for D-day

Oppenheimer1945 · Answer

2n-1=739, 2m-1=1479

(1+....739)/(1+3+....1479-1-3-5-....-739)

=370^2/(740^2-370^2)=1/(2^2-1)

Lewy · Answer

nth term is given as 2n -1
there4 in 2n -1 =739, n =370
Sum of the terms in the series is given by n^2, 
hence upper fraction will be 370^2
lower fraction contains terms between 741 thru 1479
last term, nth term is 2n - 1 = 1479, n = 740

The sum of lower terms is 740^2 - 370^2 which is the same as (a^2 - b^2) = (a + b) (a - b)
There4 the whole expression can be written as 370^2 /(740^2 - 370^2) 
(370 *370) / (740+370)(740 - 370) = 1/3

For positive integers 'n', it is known that 1 + 3 + 5 + ... + (2*n - 1

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For positive integers 'n', it is known that 1 + 3 + 5 + ... + (2*n - 1

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