Question

Write the expansion of (1 – x):
om
Hins​

Answer 1

Answer:

If n is positive integer or positive rational number, we can have the binomial expansion of (1+x)−n as an infinite series as follows:

(1+x)−n=1+−n1.x1+(−n)(−n−1)1×2.x2+(−n)(−n−1)(−n−2)1×2×3.x3+...+(−n)(−n−1)(−n−2)...(−n−r+1)1×2...×r.xr+...∞

This series converges to some real/complex value, for each x with |x|<1

For example (1+x)−1=1+−11.x1+(−1)(−1−1)1×2.x2+(−1)(−1−1)(−1−2)1×2×3.x3+...+...∞

=1−x1+x2−x3+...∞

(1+x)−2=1+−21.x1+(−2)(−2−1)1×2.x2+(−2)(−2−1)(−2−2)1×2×3.x3+...+...∞

=1−2x1+3x2−4x3+...∞

(1+x)−12=1+−121.x1+(−12)(−12−1)1×2.x2+...∞

=1−12x1+1×32×4x2−1×3×52×4×6x3+...∞

Actually these are Taylor series of the given function about the point zero.

Answer 2

Step-by-step explanation:

We can write down the binomial expansion of (1+x)n(1+x)n as

1+n1!x+n(n−1)2!x2+n(n−1)(n−2)3!x3+...

1+n1!x+n(n−1)2!x2+n(n−1)(n−2)3!x3+...

This is true for all real values of nn, although there are conditions on xx.

If nn is a positive integer, the expansion terminates, while if nn is negative or not an integer (or both), we have an infinite series that is valid if and only if ||x||<1|x|<1.

We can now compare this with the series we are given. The increasing powers of 1313 strongly suggest that x=13x=13.

What can we now say about nn? If we look at the x2x2 term, we need 12×32=n(n−1)12×32=n(n−1). The only ways this can happen are if n=−12n=−12 or n=32n=32.

The x3x3 term tells us −12×32×52=n(n−1)(n−2)−12×32×52=n(n−1)(n−2), which must mean n=−12n=−12.

We can easily check that this provides the other terms of the series successfully.

Thus x=13,n=−12x=13,n=−12.