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Difficulty:
45%
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Question Stats:
70% (02:10) correct
30%
(02:49)
wrong
based on 169
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Let \(n\) be a positive integer such that \((n + 1)!(n + 2)! + (n + 2)!(n + 3)! = (n + 1)!(n + 2)!(pn^2 + qn + r)\), then what is the value of \(p + q + r\)?
(A) 10
(B) 11
(C) 12
(D) 13
(E) 14
(A) 10
(B) 11
(C) 12
(D) 13
(E) 14
DeepGOO225
Joined: 10 Nov 2025
Last visit: 01 Sep 2026
Posts: 29
Own Kudos:
Given Kudos: 3
Location: India
Schools: ESSEC MiM '27 ISB HEC MiM '27
GPA: 3.22
06 Aug 2026, 02:26
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ANS should be 13
1. Factor the common factorial immediately.
You see
(n+1)!(n+2)!+(n+2)!(n+3)!
Your eyes should instantly notice the common factor:
(n+2)!
2. Rewrite the larger factorial.
Always reduce larger factorials to match the smaller one.
Here,
(n+3)!=(n+3)(n+2)(n+1)!
This lets you factor out (n+1)! too.
3. Expand only once.
Compute
(n+2)(n+3)=n 2+5n+6
Then add the extra 1:
n 2+5n+7.
4. Match coefficients.
Since
pn 2 +qn+r=n 2 +5n+7,
read off:
p=1
q=5
r=7
No need for lengthy algebra.
1. Factor the common factorial immediately.
You see
(n+1)!(n+2)!+(n+2)!(n+3)!
Your eyes should instantly notice the common factor:
(n+2)!
2. Rewrite the larger factorial.
Always reduce larger factorials to match the smaller one.
Here,
(n+3)!=(n+3)(n+2)(n+1)!
This lets you factor out (n+1)! too.
3. Expand only once.
Compute
(n+2)(n+3)=n 2+5n+6
Then add the extra 1:
n 2+5n+7.
4. Match coefficients.
Since
pn 2 +qn+r=n 2 +5n+7,
read off:
p=1
q=5
r=7
No need for lengthy algebra.
06 Aug 2026, 02:29
Kudos
Bookmarks
(n+1)!(n+2)!(n2+5n+7)=(n+1)!(n+2)!(pn2+qn+r)
therefore, 1+5+7=13
Why 12 ?
therefore, 1+5+7=13
Why 12 ?
