Question

Find a positive value of m for which the coefficient of x2 in expansion of (1+x)m is 6

Answer 1

AnswEr:

We know that the coefficient of $\bf{x}^{r}$ in $\bf{(1+x)^n}$ is $\bf{^nC_r}$. Therefore, coefficient of x² in $\bf{(1+x)^m}$ is $\bf{^mC_2}$.

• It is given that the coefficient of x² in $\bf{(1+x)^m}$ is 6.

$\therefore \tt {}^{m} C_2 = 6 \\ \\ \rightarrow \tt \frac{m(m - 1)}{2!} = 6 \\ \\ \rightarrow \tt {m}^{2} - m = 12 \\ \\ \rightarrow \tt {m}^{2} - m - 12 = 0 \\ \\ \tt \rightarrow(m - 4)(m + 3) = 0 \\ \\ \tt \rightarrow \: m - 4 = 0 \\ \\ \implies \tt \: m = 4$

Since, m + 3 is not equal to 0.

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• Coefficient of (r+1)th term in the binomial expansion of $\bf{(1+x)^n}$ is $\bf{^nC_r}$.

• Coefficient of $\bf{x}^{r}$ in the binomial expansion of $\bf{(1+x)^n}$ is $\bf{^nC_r}$.