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Give Brief solution of the question in the attachment.
one month ago

from binomial theorem,
(1+x)^100= 100C0 + 100C1*x + 100C2x^2 + 100C3*x^3 + …........
in the above identity, put x= 1, w, w^2 successively (where w and w^2 are the complex cube roots of unity, as we know that roots of z^3= 1 are 1, w, w^2)
so, 2^100= 100C0 + 100C1 + 100C2 + 100C3 + …........
and (1+w)^100= 100C0 + 100C1*w + 100C2w^2 + 100C3*w^3 + …........
and (1+w^2)^100= 100C0 + 100C1*w^2 + 100C2w^4 + 100C3*w^6 + …........
adding these 3 eqns,
2^100 + (1+w)^100 + (1+w^2)^100= 3*100C0 + 100C1*(1+w+w^2) + 100C2*(1+w^2+w^4) + 100C3*(1+1+1) + ….....
or 2^100 + (1+w)^100 + (1+w^2)^100= 3*100C0 + 0 + 0 + 3*100C3 + 0 + …..............(because w^3= 1 and 1+w+w^2= 0)
or  2^100 + (1+w)^100 + (1+w^2)^100= 3(100C0 + 100C3 + …......)
so, 100C0 + 100C3 + ….....= (2^100 + (1+w)^100 + (1+w^2)^100)/3
so, we just need to find value of (1+w)^100 + (1+w^2)^100= ( – w^2)^100 + ( – w)^100 [using 1+w+w^2= 0]
= w^200 + w^100= w^100(1+w^100)= w^100(1+(w^3)^33*w)= w^100(1+w)= w^100( – w^2)= – w^102= – (w^3)^34= – 1
so, final ans is (2^100 – 1)/3
kindly approve :=)
one month ago
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