What will be the constant term in the expansion of $$\left(x + \frac{1}{x}\right)^6$$ ?

Explanation

Constant term in $$\left(x + \frac{1}{x}\right)^6$$ – From the binomial theorem, $${}^6C_0(x^6) + {}^6C_1(x^5)\left(\frac{1}{x}\right) + {}^6C_2(x^4)\left(\frac{1}{x}\right)^2 + {}^6C_3(x^3)\cdot\left(\frac{1}{x}\right) + {}^6C_4(x^2)\cdot\left(\frac{1}{x}\right)^4 + {}^6C_5(x)\cdot\left(\frac{1}{x}\right)^5 + {}^6C_6\left(\frac{1}{x}\right)^6$$ = $$x^6 + 6.x^5.\frac{1}{x} + 15.x^4\frac{1}{x^2} + 20.x^3\frac{1}{x^3} + 15.x^2\frac{1}{x^4} + 6.x.\frac{1}{x^5} + \frac{1}{x^6}$$ = $$x^6 + 6x^4 + 15x^2 + 20 + \frac{15}{x^2} + \frac{6}{x^4} + \frac{1}{x^6}$$ Therefore, the constant term is 20.