Let ,,, be the coefficients of x^7,x^5,x^3,x respectively in the expansion of (x+x^3-1)^5+…

Question

Let $\alpha,\beta,\gamma,\delta$ be the coefficients of $x^7,x^5,x^3,x$ respectively in the expansion of $\left(x+\sqrt{x^3-1}\right)^5+\left(x-\sqrt{x^3-1}\right)^5$, $x>1$. If $u$ and $v$ satisfy the equations

$$\alpha u+\beta v=18,\qquad \gamma u+\delta v=20$$

then $u+v$ is equal to :

Accepted Answer

By the binomial theorem for $(a+b)^5+(a-b)^5=2\left[\binom50a^5+\binom52a^3b^2+\binom54ab^4\right]$ with $a=x,\,b^2=x^3-1$:

$$\left(x+\sqrt{x^3-1}\right)^5+\left(x-\sqrt{x^3-1}\right)^5 = 2\left[x^5+10x^3(x^3-1)+5x(x^3-1)^2\right]$$

$$=2\left(5x^7+10x^6+x^5-10x^4-10x^3+5x\right)$$

So $\alpha=10$ (coeff of $x^7$), $\beta=2$ (coeff of $x^5$), $\gamma=-20$ (coeff of $x^3$), $\delta=10$ (coeff of $x$).

The equations become: $10u+2v=18\Rightarrow5u+v=9$, and $-20u+10v=20\Rightarrow-2u+v=2$.

Subtracting: $7u=7\Rightarrow u=1$, then $v=9-5=4$.

$u+v=1+4=5$

Solutions · 1