Question

rth term in the expansion of $(a+2x)^n$ is

• A. $\frac{n(n+1)....(n-r+1)}{r!} a^{n-r+1} (2x)^r$
• B. $\frac{n(n+1)....(n-r)}{(r+1)!} a^{n-r} x^r$
• C. $\frac{n(n-1)....(n+r)}{(r-1)!} a^{n-r} x^r$
• D. $\frac{n(n-1)...(n-r+2)}{(r-1)!} a^{n-r+1} (2x)^{r-1}$

Answer 1

Answer: (D)

$\frac{n(n-1)...(n-r+2)}{(r-1)!} a^{n-r+1} (2x)^{r-1}$

Answer 2

The general term in the binomial expansion of $(X+Y)^n$ is given by $T_{k+1} = \binom{n}{k} X^{n-k} Y^k$.

In the given problem, we have the expansion of $(a+2x)^n$. Here, $X = a$ and $Y = 2x$. We need to find the $r^{th}$ term, which means $T_r$. To find $T_r$, we set $k+1 = r$, which implies $k = r-1$.

Substitute $k = r-1$, $X = a$, and $Y = 2x$ into the general term formula: $T_r = \binom{n}{r-1} a^{n-(r-1)} (2x)^{r-1}$ $T_r = \binom{n}{r-1} a^{n-r+1} (2x)^{r-1}$

Now, let's express the binomial coefficient $\binom{n}{r-1}$ in expanded form: $\binom{n}{r-1} = \frac{n!}{(r-1)!(n-(r-1))!} = \frac{n!}{(r-1)!(n-r+1)!}$

To match the form presented in the options, we can write $n!$ as a product of terms: $n! = n \times (n-1) \times (n-2) \times \dots \times (n-r+2) \times (n-r+1)!$

Substitute this back into the expression for $\binom{n}{r-1}$: $\binom{n}{r-1} = \frac{n(n-1)(n-2)\dots(n-r+2)(n-r+1)!}{(r-1)!(n-r+1)!}$ $\binom{n}{r-1} = \frac{n(n-1)(n-2)\dots(n-r+2)}{(r-1)!}$

Finally, substitute this back into the expression for $T_r$: $T_r = \frac{n(n-1)(n-2)\dots(n-r+2)}{(r-1)!} a^{n-r+1} (2x)^{r-1}$

Comparing this result with the given options, we find that it matches the fourth option.