Binomial Theorem. A little bit challenging

1) Obtain the first four terms in the expansion of (1+x+x^2)^10.Hence find the approximate value of 1.0101^10, giving your answer to 3 decimal place.

Please show me full workings and explanations. Please don't reply with "Google it" or anything. I really want to learn. Anyone can teach me out there?

(1+x+x^2)^10 = 1 + 10(x + x^2) + 45(x + x^2)^2 + 120(x + x^2)^3 + ...
for x = 0.01, 1+x+x^2 = 1.0101
so 1.0101^10 ~= 1 + 10(0.0101) + 45(0.0101)^2 + 120(0.0101)^3 = 1.106

(1+x+x^2)¹⁰ =1 +C(10,1)(x+x²)+C(10,2)(x+x²)² +C(10,3)(x+x²)³+...................
1.0101 =1+.01+.0001=1+.01+(.01)²
I.e x= 0.01→1+x+x²=1.0101
C(10,1)=10
C(10,2)= 45
C(10,3) =120
C(10,4) =210
C(10,5) =252
C(10,6) =210
C(10,7) =120
C(10,8) =45
C(10,9) =10
C(10,10) =1
(1+x+x^2)¹⁰ =1 +10(x+x²)+45(x+x²)² +120(x+x²)³+210(x+x²)⁴+252(x+x²)⁵+210(x+…‡
10(x+x²) =10(0.01+0.001) =0.1010
45(x+x²)² =45(.01+0.0001)²=45(.0001+0.00000001+0.0…‡
120(x+x²)³=120(0.01+0.0001) =120(0.000001+ 0.00000003+0.0000000003+0.000000000001)
=120(0.000001030301) =0.00012363612
Hence (1.0101)¹⁰=1+0.1010+0.00459045+.00012363…‡ =1.105714083612=1.106 upto three decimal places