Complex number 1!

Algebra Level 5

If $x = 2+5i$ and $\large 2(\frac{1}{1!9!} +\frac{1}{3!7!}) + \frac{1}{5!5!} = \frac{2^a}{b!}$ , then the value of $(x^3 -5x^2 +33x -19)$ is equal to

a-b

a

None of These

a+b

b

1 solution

Tanishq Varshney
May 30, 2015

$(x-2)^{2}=-25$

$x^{2}-4x+29=0$

Also

$\large{\frac{1}{10!}(2(\binom{10}{1}+\binom{10}{3})+\binom{10}{5})}$

using $\binom{n}{1}+\binom{n}{3}+\binom{n}{5}+...+\binom{n}{n-1}=2^{n-1}$

so $\large{\frac{1}{10!}(2(\binom{10}{1}+\binom{10}{3})+\binom{10}{5})=\frac{2^{9}}{10!}}$

so $a=9$ and $b=10$

Now we have the expression $x^{3}-5x^{2}+33x-19$

which can be rearranged as $x(x^{2}-4x+29)-(x^{2}-4x+29)+10$

$10=b$

nicely done, upvoted $\ddot \smile$ !

Nishant Rai - 6 years ago

thanx , can u post a solution for your problem 2 and 3 on complex number

Tanishq Varshney - 6 years ago