Let the coefficients of three consecutive terms T_r, T_r+1 and T_r+2 in the binomial expan…

Question

Let the coefficients of three consecutive terms $T_r$, $T_{r+1}$ and $T_{r+2}$ in the binomial expansion of $(a + b)^{12}$ be in a G.P. and let $p$ be the number of all possible values of $r$. Let $q$ be the sum of all rational terms in the...

Accepted Answer

$\begin{gathered} (a+b{)}^{\frac{1}{2}} \ {T}_{r},{T}_{r+1},{T}_{r+2}	o GP \ \frac{{T}_{r+1}}{{T}_{r}}=\frac{{T}_{r+2}}{{T}_{r+1}} \ \frac{{}^{12}{C}_{r}}{{}^{12}{C}_{r-1}}=\frac{{}^{12}{C}_{r+1}}{{}^{12}{C}_{r}} \ \frac{12-r+1}{r}=\frac{12-(r+1)+1}{r+1} \ (13-r)(r+1)=(12-r)(r) \ -r+12r+13=12r-{r}^{2} \ 13=0 \end{gathered}$

No value of r possible

So $P=0$

${({3}^{\frac{1}{4}}+{4}^{\frac{1}{3}})}^{12}=\sum {}^{12}{C}_{r}{({3}^{\frac{1}{4}})}^{12-r}{({4}^{\frac{1}{3}})}^{r}$

Exponent of $({3}^{\frac{1}{4}})$ exponent of $({4}^{\frac{1}{3}})$ term

$\begin{gathered} 12 \quad  0 \quad  27 \ 0 \quad  12 \quad  256 \ q=27+256=283 \quad  \quad  \ p+q=0+283=283 \quad  \quad  \end{gathered}$

Solutions · 1