(a+b+c)^1 = 2a
(a+b+c)^3 + (a-b-c)^3 = 2a(a^2 + 3(b+c)^2)
(a+b+c)^5 + (a-b-c)^5 = 2a(a^4 + 10a^2(b+c)^2 + 5(b+c)^4)
So, for k=n, there are
1: 1
3: 1+3
5: 1+3+5
these are the odd squares.
The answer must be 2017^2
(a+b+cc)^(2017)+(a-b-c)^(2017)
When simplified how many terms would that be?
can we try this? i saw it in IMO past question I couldn't do it
Sir obleck I search for that book you said ,but no luck for me ....I only have 2014 past question but it hasn't been helpful so far...I need to understand the ideal and fast thinking to question like this.....
2 answers
Just the right person I was looking for thanks man
1 answer