Question

If $a+b+c+d=0$, shew that $\frac{a^5+b^5+c^5+d^5}{4}=\frac{a^3+b^3+c^3+d^3}{3}\cdot \frac{a^2+b^2+c^2+d^2}{2}$.

Answer 1

$(1+ax)(1+bx)(1+cx)(1+dx)=1+q{x}^2+r{x}^3+s{x}^4\because \sum a=0$

$q=\sum ab,r=\sum abc,s=abcd$

Taking the logarithms and equating the co efficients of $x^n$, we have $\frac{{(-1)}^{n-1}}{n}(a^n+b^n+c^n+d^n)$

$=$ co efficient of $x^n$ in the expansion of $\log (1+q{x}^2+r{x}^3+s{x}^4)$

$=$ co efficient of $x^n$ in $(q{x}^2+r{x}^3+s{x}^4)$

$=-\frac{1}{2}{(q{x}^2+r{x}^3+s{x}^4)}^2+....$

By putting $n=2,3,4,5$ we obtain

$-\frac{a^2+b^2+c^2+d^2}{2}=q$

$\frac{a^3+b^3+c^3+d^3}{3}=r$

$\frac{a^5+b^5+c^5+d^5}{5}=-qr$

$\therefore \frac{\sum a^5}{5}=\frac{\sum a^3}{3}.\frac{\sum a^2}{2}$