Super awesome simplification.

Algebra Level 3

Find the value of - ( 4 ) . ( 1 + 5 4 ) . ( 1 + 7 9 ) . . . . . . . . . . . . ( 1 + 4027 4052169 ) . ( 1 + 4029 4056196 ) (4).(1+\frac{5}{4}).(1+\frac{7}{9})............(1+\frac{4027}{4052169}).(1+\frac{4029}{4056196})


The answer is 4060225.

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3 solutions

The product becomes n = 1 2014 ( 1 + 2 n + 1 n 2 ) = n = 1 2014 ( n + 1 ) 2 n 2 = 201 5 2 = 4060225 \displaystyle \prod\limits_{n=1}^{2014} \left(1 + \frac{2n+1}{n^{2}}\right) = \prod\limits_{n=1}^{2014} \frac{(n+1)^{2}}{n^{2}} = 2015^{2} = \boxed{4060225} ~~~

Excellent and elegant :)

Krishna Ar - 6 years, 10 months ago

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What? Level 3 ???? Sorry Krish but it was not of that standard!!! Looking forward to your 1000th follower! And I am looking forward to my 100th follower!! :( :P

Kartik Sharma - 6 years, 10 months ago

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@Kartik Sharma -I am a huge fan of your brilliance..this problem's now back to lvl 2..happy??/

Krishna Ar - 6 years, 10 months ago
Sachin Mittal
Aug 1, 2014

Try to observe the pattern---

(4).(9/4).(16/9).......(4056196/4052169).(4060225/4056196)

Here,you can observe that---numerator of the terms are same as the denominator of the next term to them,except the last term.So,

(1).(1/1).(1/1)....(1/1)(4060225/1)=4060225

Nikhil Tandon
Aug 18, 2014

Just observe the pattern. Everything else cancels out except the last term

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