Binomial Theorem

[krtxmrtz]

And, how to find the coefficient of x in the expansion of
[x-(2/x^2)]^16?

Thanks again.

The expansion of (a + b)ⁿ has n+1 terms and the j-th term is:

[n! / ((j - 1)!*(n - j + 1)!)] a^{n - j + 1}b^{j - 1}

and the sum of the powers of a and b is always n:
n - j + 1 + (j - 1) = n

In your specific case, the j-th term is:

[16! / ((j - 1)!*(17 - j)!)] x^{16 - j + 1}(-2/x²)^{j - 1} = [16! / ((j - 1)!*(17 - j)!)] x^{17 - j}(-2*x^-2)^{j - 1} = [16! / ((j - 1)!*(17 - j)!)] x^{17 - j}(-2)^{j - 1}*x^{-2(j - 1)}

so you're aiming at a j value such that the power of x is 1,

17 - j -2(j - 1) = 1 wherefore j = 6, so finally the term you want is:

[16! / ((6 - 1)!*(17 - 6)!)] x^{17 - 6}(-2)^{6 - 1}*x^{-2(6 - 1)} = [16! / (5! * 11!)]x¹¹*(-2)⁵x^-10 = -4362*32 x = -139776 x

And to be strict, the coefficient is just -139776