The coefficient of x^18 in the expansion of (1+x^2+2x)^15 is: — Solution | JEE Mathematics

NVAEasyJEE 2023General Term

The coefficient of $x^{18}$ in the expansion of $(1+x^{2}+2x)^{15}$ is:

Answer

Correct answer:5005

Step-by-step solution

Standard Method

Given: We need the coefficient of $x^{18}$ in the expansion of $(1+x^{2}+2x)^{15}$ .

Find: The numerical value of the required coefficient.

Observe that

1+x^{2}+2x = (x+1)^{2}

Therefore,

(1+x^{2}+2x)^{15} = \left((x+1)^{2}\right)^{15} = (x+1)^{30}

Now use the general term in the binomial expansion of $(1+x)^{30}$ :

T_{r+1}=\binom{30}{r}x^{r}

For the coefficient of $x^{18}$ , take $r=18$ . Hence the coefficient is

\binom{30}{18}=\binom{30}{12}=86493225

Therefore, the coefficient of $x^{18}$ is $86493225$ .

About the extracted solution

The solution works with the expression $\left(x^{4}-\frac{1}{x^{3}}\right)^{15}$ and concludes a different coefficient. That working does not match the given question text, so it cannot be used as the solution for this question.

Common mistakes

Expanding $1+x^{2}+2x$ incorrectly. Since $1+x^{2}+2x=(x+1)^{2}$ , first simplify the bracket before raising it to the power. Missing this identity leads to the wrong binomial form.
Using the wrong exponent after simplification. From $((x+1)^{2})^{15}$ , the result is $(x+1)^{30}$ , not $(x+1)^{15}$ . Always multiply the powers correctly.
Choosing the wrong binomial term. In $(1+x)^{30}$ , the coefficient of $x^{18}$ is $\binom{30}{18}$ . Do not use $\binom{15}{6}$ or values taken from an unrelated expression.

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