Question

If the coefficients of $\sum_{k = 0}^{10}{20C_{k} =}$, $2^{19} + \frac{1}{2}^{20}C_{10}$ and $2^{19}$terms in the expansion of $20C_{10}$are in A.P., then.

• A. $\frac{2n - 1}{2}$
• B. $\frac{1}{2}n - 1$
• C. $n - 1$
• D. None of these

Answer 1

Answer: (B)

$\frac{1}{2}n - 1$

Answer 2

Coefficient of $(r + 4)osa$and $a_{1},a_{2},a_{3},a_{4}$ terms in expansion of $a_{1} =^{n}C_{r},a_{2} =^{n}C_{r + 1},a_{3} =^{n}C_{r + 2,}a_{4} =^{n}C_{r + 3}$ are $\frac{a_{1}}{a_{1} + a_{2}} + \frac{a_{3}}{a_{3} + a_{4}} = \frac{nC_{r}}{nC_{r} +^{n}C_{r + 1}}$.

Then $= \frac{r + 1}{n + 1} + \frac{r + 3}{n + 1} = \frac{2(r + 2)}{n + 1}$

$= 2\frac{nC_{r + 1}}{n + 1C_{r + 2}} = 2\frac{nC_{r + 1}}{nC_{r - 1} +^{n}C_{r + 2}} = \frac{2a_{2}}{a_{2} + a_{3}}$

Trick : Let p = 1, hence $T_{r + 1} =^{642}C_{r}(5^{1/2})^{642 - r}.(7^{1/6})^{r}$and $= \frac{642}{6} = 107$are in A.P.

$(2 + \sqrt{2})^{4} = (\sqrt{2})^{4}(\sqrt{2} + 1)^{4}$

$4\lbrack^{4} ⥂ C_{0} +^{4} ⥂ C_{1}(\sqrt{2}) +^{4} ⥂ C_{2}(\sqrt{2})^{2} +^{4} ⥂ C_{3}(\sqrt{2})^{3} +^{4} ⥂ C_{4}(\sqrt{2})^{4}\rbrack$

which is given by (2).