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Evaluate (1^4+2007^4+2008^4)/(1^2+2007^2+2008^2) | IOQM

JEE Maths question with a full step-by-step solution.

Question
Find the value of
14+20074+2008412+20072+20082,\quad \frac{1^4 + 2007^4 + 2008^4}{1^2 + 2007^2 + 2008^2}, \qquad\qquad
Solution
Answer: 4030057
For all real a,ba, b,
a4+b4+(a+b)4=12[a2+b2+(a+b)2]2a^4 + b^4 + (a+b)^4 = \frac{1}{2}\big[a^2 + b^2 + (a+b)^2\big]^2
Since 2008=1+20072008 = 1 + 2007, take a=1a = 1, b=2007b = 2007, a+b=2008a + b = 2008.
14+20074+2008412+20072+20082=12[12+20072+20082]212+20072+20082=12(1+20072+20082)\frac{1^4 + 2007^4 + 2008^4}{1^2 + 2007^2 + 2008^2} = \frac{\tfrac{1}{2}\big[1^2 + 2007^2 + 2008^2\big]^2}{1^2 + 2007^2 + 2008^2} = \frac{1}{2}\big(1 + 2007^2 + 2008^2\big)
With 20072=40280492007^2 = 4028049 and 20082=40320642008^2 = 4032064,
=12(1+4028049+4032064)=12(8060114)=4030057= \frac{1}{2}(1 + 4028049 + 4032064) = \frac{1}{2}(8060114) = 4030057
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