First slide

Binomial theorem for positive integral Index

Question

If in the expansion of (a - 2b)ⁿ ,the sum of 5^th and 6^th terms is 0, then the values of a/b =

Moderate

A

$\frac{n - 4}{5}$
B

$\frac{2 (n - 4)}{5}$
C

$\frac{5}{n - 4}$
D

$\frac{5}{2 (n - 4)}$

Solution

$T_{5} =^{n} C_{4} a^{n - 4} (- 2 b)^{4 \{G e n e r a l t e r m = T_{r + 1} = n_{c_{r}} x^{n - r} a^{r}\}}$
$\begin{array}{l} and T_{6} =^{n} C_{5} a^{n - 5} (- 2 b)^{5} \\ As T_{5} + T_{6} = 0, we get \\ ^{n} C_{4} 2^{4} a^{n - 4} b^{4} =^{n} C_{5} 2^{5} a^{n - 5} b^{5} \\ or \frac{a^{n - 4} b^{4}}{a^{n - 5} b^{5}} = \frac{n! 2^{5}}{5! (n - 5)!} \cdot \frac{4! (n - 4)!}{n! 2^{4}} \\ or \frac{a}{b} = \frac{2 (n - 4)}{5} \end{array}$

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