Question

If the coefficients of $a^{r-1},a^r$ and $a^{r+1}$ in the expansion of $(1+a{)}^n$ are in arithmetic progression, then $n^2-n(4r+1)+4r^2-2$ is equal to

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (A)

The $(r+1{)}^{\text{th }}$ term in the expansion is ${}^n{C}_r{a}^r$. Thus, it can be seen that $a^{\prime }$ occurs in the $(r+1{)}^{\text{th }}$ term, and its coefficient is ${}^n{C}_r$. Hence, the coefficients of $a^{r-1},a^r$ and $a^{r+1}$ are ${}^n{C}_{r-1},{}^n{C}_r$ and ${}^n{C}_{r+1}$, respectively.
Since, these coefficients are in arithmetic progression. So, we have ${}^n{C}_{r-1}+{}^n{C}_{r+1}=2.{}^n{C}_r$. This gives
$\frac{n!}{(r-1)!(n-r+1)!}+\frac{n!}{(r+1)!(n-r-1)!}=2\times \frac{n!}{r!(n-r)!}$
i.e., $\frac{1}{(r-1)!(n-r+1)(n-r)(n-r-1)!}$
$+\frac{1}{(r+1)(r)(r-1)!(n-r-1)!}$
$=2\times \frac{1}{r(r-1)!(n-r)(n-r-1)!}$
or $\frac{1}{(r-1)!(n-r-1)!}\left[\frac{1}{(n-r)(n-r+1)}+\frac{1}{(r+1)(r)}\right]$
$=2\times \frac{1}{(r-1)!(n-r-1)![r(n-r)]}$
i.e., $\frac{1}{(n-r+1)(n-r)}+\frac{1}{r(r+1)}=\frac{2}{r(n-r)}$,
or $\frac{r(r+1)+(n-r)(n-r+1)}{(n-r)(n-r+1)r(r+1)}=\frac{2}{r(n-r)}$
or $r(r+1)+(n-r)(n-r+1)=2(r+1)(n-r+1)$
or $r^2+r+n^2-nr+n-nr+r^2-r=2\left(nr-r^2+r+n-r+1\right)$
or $n^2-4nr-n+4r^2-2=0$
i.e., $n^2-n(4r+1)+4r^2-2=0$