Question

The Sum of the coefficients in $(x+2y+z{)}^{10}$ is:

• A. 2¹⁰
• B. 3¹⁰
• C. 1
• D. 4¹⁰

Answer 1

Answer: (D)

In Binomial expansion,

$(x+y{)}^n=n_{c_0}{x}^n+n_{c_1}{x}^{n-1}y+n_{c_2}{x}^{n-2}{y}^2+....+n_{c_n}{y}^n$

To get the sum of the coefficients, put $x=y=1$

$\Rightarrow (1+1{)}^n=n_{c_0}(1{)}^n+n_{c_1}\left(1{)}^{n-1}\right(1)+n_{c_2}(1{)}^{n-2}{1}^2+....+n_{c_n}(1{)}^n$

$\Rightarrow 2^n=n_{c_0}+n_{c_1}+n_{c_2}+....+n_{c_n}$

Now, To get the sum of the coefficients, in $(x+2y+z{)}^{10}$

put $x=y=z=1$

$=(1+2(1)+1{)}^{10}$

$=4^{10}$

Hence, Option D is correct.