Single Choice

The sum of the coefficients of even powers of $$x$$ in the expansion of $$ (1+x+x^2+x^3)^5 $$ is

A$$512$$
Correct Answer
B$$256$$
C$$128$$
D$$64$$

Solution

Let $$f(x)=(1+x+x^2+x^3)\:^5$$
$$=a_{0}+a_{1}x+a_{2}x^2+a_{3}x^3...$$ ...(i)
$$f(1)=a_{0}+a_{1}+a_{2}...$$ ...(ii)
$$f(-1)=a_{0}-a_{1}+a_{2}-a_{3}...$$ ...(iii)
Adding (ii) and (iii), we get
$$f(1)+f(-1)=2[a_{0}+a_{2}+a_{4}...]$$
$$(1+1+1+1)^{5}+(1-1+1-1)^{5}=2[a_{0}+a_{2}+a_{4}...]$$
$$4^{5}=2[a_{0}+a_{2}+a_{4}...]$$
$$2^{10}=2[a_{0}+a_{2}+a_{4}...]$$
$$2^9=a_{0}+a_{2}+a_{4}...$$
$$512=a_{0}+a_{2}+a_{4}...$$
Hence sum of the coefficients of even powers of $$x$$ is $$512$$.

SIMILAR QUESTIONS

Binomial Theorem

Sum of the coefficients of $$ (1+x)^n $$ is always a

Binomial Theorem

Sum of the coefficients of $$ (1 - x)^{25} $$ is

Binomial Theorem

Let $$(x + 10)^{50} + (x - 10)^{50} = a_0 + a_1x + a_2x^2 + .... + a_{50} x^{50}$$, for all $$x \in R$$ then $$\dfrac{a_2}{a_0}$$ is equal to:-

Binomial Theorem

The sum of the co-efficients of all odd degree terms in the expansion of $$\left(x + \sqrt {x^{3} - 1}\right)^{5} + (x - \sqrt {x^{3} - 1})^{5}, (x > 1)$$ is

Binomial Theorem

Find the sum of coefficients in the expansion of $$\left(1-\dfrac 2x+\dfrac {4}{x^2}\right)^n$$ given that the number of terms are $$28$$

Binomial Theorem

The value of $$\displaystyle \sum_{r = 1}^{15} r^{2} \left (\dfrac {^{15}C_{r}}{^{15}C_{r - 1}}\right )$$ is equal to:

Binomial Theorem

The sum of coefficient of integral powers of $$x$$ in the binomial expansion of $$(1-2\sqrt x)^{50}$$ is :

Binomial Theorem

If the sum of odd numbered terms and the sum of even numbered terms in the expansion of $$(x + a)^{n}$$ are $$A$$ and $$B$$ respectively, then the value of $$(x^{2} - a^{2})^{n}$$ is

Binomial Theorem

Let $$X = (^{10}C_{1})^{2} + (^{10}C_{2})^{2} + 3(^{10}C_{3})^{2} + .... + 10 (^{10}C_{10})^{2}$$, where $$^{10}C_{r}, r\epsilon \left \{1, 2, ..., 10\right \}$$ denote binomial coefficients. Then, the value of $$\dfrac {1}{1430}X$$ is

Binomial Theorem

The sum of the coefficients of even powers of $$x$$ in the expansion of $$ (1+x+x^2+x^3)^5 $$ is

Contact Details