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Bunuel
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If f(m!) = (m!)^2, for all positive integers m, then what will f(15!) 17 Jan 2022, 03:15
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E
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If \(f(m!) = (m!)^2\), for all positive integers m, then what will \(f(15!) - f(14!)\) equal to?


A. \(f(1!)\)

B. \(f(14!)*f(4!)*7\)

C. \(f(14!)*16*7\)

D. \(f(15!)\)

E. \(f(14!)*16*14\)



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E
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happyposeidon835
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If f(m!) = (m!)^2, for all positive integers m, then what will f(15!) 23 Jan 2022, 17:11
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f(15!)−f(14!)

=> 15!*15! - 14!*14!

=> 15*14!*15*14! - 14!*14!

=> 14!*14!*(15*15 - 1)

=> f(14!) *224

=> f(14!) *16 *14

So, option E
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If f(m!) = (m!)^2, for all positive integers m, then what will f(15!) 24 Jan 2022, 21:36
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Given f(m!) = \(\left(m!\right)^2\).

f(15!) - f(14!) = \(\left(15!\right)^2-\ \left(14!\right)^2\)

This can be rewritten as :\( \left(15!+14!\right)\left(15!-14!\right)\)

= Taking 14! element from both the terms we have :

\(14!\cdot14!\cdot\left(15+1\right)\left(15-1\right)\)

= 14!*14!*(14)*(16)

= f(14)*(14)*(16)
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If f(m!) = (m!)^2, for all positive integers m, then what will f(15!) 03 Dec 2024, 19:24
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\(f(m!) = (m!)^2\) and we need to find \(f(15!) - f(14!)\)

=> To find the value of f(15!) we need to replace m with 15 in \(f(m!) = (m!)^2\)
=> \(f(15!) = (15!)^2\)

=> \(f(15!) - f(14!)\) = \((15!)^2 - (14!)^2\)
= \((15*14!)^2 - (14!)^2 \)
= \( (14!)^2 * ( 15^2 - 1) \)
= \( (14!)^2 * (225 - 1) \)
= \( (14!)^2 * 224 \)
= \( (14!)^2 * 16 * 14 \)
= \( f(14!) * 16 * 14\)

So, Answer will be E
Hope it helps!

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If f(m!) = (m!)^2, for all positive integers m, then what will f(15!) 04 Dec 2024, 08:36
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why do you need to take out 14! twice why cant it just be once like 14!(16)(14)?
GmatPoint
Given f(m!) = \(\left(m!\right)^2\).

f(15!) - f(14!) = \(\left(15!\right)^2-\ \left(14!\right)^2\)

This can be rewritten as :\( \left(15!+14!\right)\left(15!-14!\right)\)

= Taking 14! element from both the terms we have :

\(14!\cdot14!\cdot\left(15+1\right)\left(15-1\right)\)

= 14!*14!*(14)*(16)

= f(14)*(14)*(16)
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If f(m!) = (m!)^2, for all positive integers m, then what will f(15!) 04 Dec 2024, 08:54
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kiasha
why do you need to take out 14! twice why cant it just be once like 14!(16)(14)?
GmatPoint
Given f(m!) = \(\left(m!\right)^2\).

f(15!) - f(14!) = \(\left(15!\right)^2-\ \left(14!\right)^2\)

This can be rewritten as :\( \left(15!+14!\right)\left(15!-14!\right)\)

= Taking 14! element from both the terms we have :

\(14!\cdot14!\cdot\left(15+1\right)\left(15-1\right)\)

= 14!*14!*(14)*(16)

= f(14)*(14)*(16)
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Because 14! is factored once from 15! + 14! and once from 15! - 14!.
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