Prove (1 + x)^n + (1 - x)^n < 2^n by using Binomial Theorem

Question

Hi my boss asked me to resolve this equation:

Prove (1 + x)^n + (1 - x)^n < 2^n by using Binomial Theorem -1 < x < 1 and n >= 2.

Answer 1

Expressing the left-hand side in binomial form :-

lhs[x_, n_] := Sum[(n!/(k!*(n - k)!))*x^k, {k, 0, n}] + 
     Sum[(n!/(k!*(n - k)!))*(-x)^k, {k, 0, n}]

It is evident that the result does not exceed 2^n.

Show[Plot[{lhs[x, 2], lhs[x, 3], lhs[x, 4]}, {x, -1, 1}], 
   Plot[{2^2, 2^3, 2^4}, {x, -1, 1}]]

Answer 2

This question seem to be about mathematics, not about Mathematica. However, Mathematica can help us to deal with this problem.

It can proof the binomial expansion of $(1-x)^n+(1+x)^n$:

Sum[2 Binomial[n, k] x^k, {k, 0, n, 2}]

(1 - x)^n + (1 + x)^n

We know that $x^k \le 1$ for $-1\le x \le 1$. Therefore $(1-x)^n+(1+x)^n$ is less than or equal to

Sum[2 Binomial[n, k], {k, 0, n, 2}]

2^n

Thereby $(1-x)^n+(1+x)^n \le 2^n$ for $-1\le x \le 1$.

Mathematica can help us to avoid simple mistakes in mathematical calculations.