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

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twobagels
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For positive integers 'n', it is known that 1 + 3 + 5 + ... + (2*n - 1 [#permalink] For positive integers 'n', it is known that 1 + 3 + 5 + ... + (2*n - 1   09 Feb 2022, 21:18
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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
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B
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houston1980
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Re: For positive integers 'n', it is known that 1 + 3 + 5 + ... + (2*n - 1 [#permalink] For positive integers 'n', it is known that 1 + 3 + 5 + ... + (2*n - 1   14 Jul 2022, 21:55
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twobagels wrote:
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



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.
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For positive integers 'n', it is known that 1 + 3 + 5 + ... + (2*n - 1 [#permalink] For positive integers 'n', it is known that 1 + 3 + 5 + ... + (2*n - 1   09 Feb 2022, 22:41
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twobagels wrote:
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


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

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

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

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


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Re: For positive integers 'n', it is known that 1 + 3 + 5 + ... + (2*n - 1 [#permalink] For positive integers 'n', it is known that 1 + 3 + 5 + ... + (2*n - 1   11 Mar 2023, 07:13
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twobagels wrote:
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


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
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Re: For positive integers 'n', it is known that 1 + 3 + 5 + ... + (2*n - 1 [#permalink] For positive integers 'n', it is known that 1 + 3 + 5 + ... + (2*n - 1   13 Mar 2023, 13:51
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twobagels wrote:
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


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

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

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

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

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

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

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

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

\(\frac{370}{ 1110 }\)

\(\frac{1}{3}\)

Option B
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Re: For positive integers 'n', it is known that 1 + 3 + 5 + ... + (2*n - 1 [#permalink] For positive integers 'n', it is known that 1 + 3 + 5 + ... + (2*n - 1   03 Sep 2023, 22:45
twobagels wrote:
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


Okay so the problem can be solved in <1min [Remember most patter type questions can actually be solved in under 1 min as long as you simply the ask]

1 + 3 + 5 + ... + (2*n - 1) = n^2 [given, if it wasnt given we could have solved it via taking a smaller case and seeing the results say 1+3/5+7 and 1+3+5/7+9+11-----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 [adding 1+3+... i.e. Num and subtracting Num, for ease]; Here n(denom) = 1480/2

Basically what we have now is (740/2)^2 / [((740*2)/2)^2-(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 :)
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Re: For positive integers 'n', it is known that 1 + 3 + 5 + ... + (2*n - 1 [#permalink] For positive integers 'n', it is known that 1 + 3 + 5 + ... + (2*n - 1   28 Jan 2024, 12:15
houston1980 wrote:
twobagels wrote:
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



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.


Could you epemd a bit on that "Pattern recognition" ?
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Re: For positive integers 'n', it is known that 1 + 3 + 5 + ... + (2*n - 1 [#permalink] For positive integers 'n', it is known that 1 + 3 + 5 + ... + (2*n - 1   29 Jan 2024, 07:54
Tukkebaaz wrote:
twobagels wrote:
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


Okay so the problem can be solved in <1min [Remember most patter type questions can actually be solved in under 1 min as long as you simply the ask]

1 + 3 + 5 + ... + (2*n - 1) = n^2 [given, if it wasnt given we could have solved it via taking a smaller case and seeing the results say 1+3/5+7 and 1+3+5/7+9+11-----Method 1]



Could you please elaborate Method 1. Cannot get how you ended up with those figures in the denominator. Thank you
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Re: For positive integers 'n', it is known that 1 + 3 + 5 + ... + (2*n - 1 [#permalink] For positive integers 'n', it is known that 1 + 3 + 5 + ... + (2*n - 1   29 Jan 2024, 10:00
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\(2n-1=739, 2m-1=1479\\
\\
(1+....739)/(1+3+....1479-1-3-5-....-739)\\
\\
=370^2/(740^2-370^2)=1/(2^2-1)\)
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Re: For positive integers 'n', it is known that 1 + 3 + 5 + ... + (2*n - 1 [#permalink]
29 Jan 2024, 10:00
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