we will get $x^{12}$ as
1. $(x^4)^3$ having coefficient $^3C_0=1$
2. $(x^3)^2(x^6)$ having coefficient $^3C_1=3$
3. $(x^3)(x^4)(x^5)$ having coefficient ${^3C_2} \times {^2C_1} =6$
So it is $10$
Just as we find binomial coefficients
We know the formula (a+b)n =n0Ca0bn+n1Ca1bn−1+......
then why you did not take rest terms.you have just taken 3C_{1}* (x^{3}) * (x^{4} + x^{5} +..)^{2}.
3C_{0} * (x^{3})^{0} * (x4 + x5 ..)^{3 } + 3C_{1}* (x^{3}) * (x^{4} + x^{5} +..)^{2}. + 3C_{2}* (x^{3})^{2} * (x^{4} + x^{5} +..)^{1}. + 3C_{3}* (x^{3}) ^{3 }* (x^{4} + x^{5} +..)^{0}
@Akriti I just have derived more, as I need coefficient of x^{12}
See, for getting coefficient of x^{12} we need put this formula to see in which sequence I am getting x^{12}
For that I can see x's power.
power of x^{4},x^{5},x^{6} giving x^{12} , rt?
then find it's coefficient. getting?
$(x^3 + x^4 + x^5 + x^6 + ... )^3 = (x^9) (1 + x^1 + x^3 + x^4 + x^5 + ...)^3$
From second term we need to find the coefficient of $x^3$ bcz ($x^3$ *$x^9$ = x^{12}) $(1 + x^1 + x^3 + x^4 + x^5 + ...)^3 = (\frac{1}{1-x})^3 = (1-x)^-3 = 1 + 3x + 6x^2 + 10x^3 +...$
Clearly, coefficient of $x^3 is 10$
Note:- Actually this was ans. from Sudarshana Tripathy which i thought that very easy to understand by anyone. but it was in the one of the comment in one of the ans here which may not be come under eye of everyone..so i thought that it is worth to put as answer.
See 5th row from below.
Trick to remember co-efficient of (1−x)^{−3 }is ...It is triangular number.
[x^{12}](x^{3} + x^{4} + x^{5} +x^{6} +...)^{3} = [x^{3}](1 + x + x^{2} + x^{3} +x^{4} +...)^{3 }= [x^{3}]((1-x)^{-1})^{3} = [x^{3}](1-x)^{-3}
^{ = }^{-3}C_{3}(-1)^{3} = ^{5}C_{3}(-1)^{3}(-1)^{3 }= ^{5}C_{3 }= 10
(x^3+x^4+x^5+x^6+…)^3 = 1/x^9 * (1+x+x^2+x^3+...)^3
now we have to find the coefficient of x^12/x^9 = x^3
this can be found in three ways - 3 OR 1+2 OR 1+1+1 = 3C1+3C1x2C1+3C3 =10
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