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In the binomial expansion of $(a-b)^{n},n\geq 5,$ then sum of 5^th and 6^th terms is zero, then $a/b$ equals

Option 1)
$\frac{n-5}{6}\;$

Option 2)
$\; \; \frac{n-4}{5}\;$

Option 3)
$\; \frac{5}{n-4}\;$

Option 4)
$\; \frac{6}{n-5}$

Answers (1)

best_answer

As we learnt in

General Term in the expansion of (x+a)^n -

$T_{r+1}= ^{n}c_{r}\cdot x^{n-r}\cdot a^{r}$

- wherein

Where $r\geqslant 0 \, and \, r\leqslant n$

$r= 0,1,2,----n$

Properties of Binomial Theorem -

$\dpi{120} \frac{^{n}c_{r-1}}{^{n}c_{r}}= \frac{r}{n-r+1}$ and

$^{n}c_{r}+^{n}c_{r+1}= ^{n+1}c_{r+1}$

-

In (a - b)ⁿ

$T_{5}=\ ^{n}C_{4}(a)^{4}(-b)^{4}$ and $T_{6}=\ ^{n}C_{5}(a)^{n-5}(-b)^{5}$

T₅ + T₆ = 0

$^{n}C_{4}(a)^{n-4}(b)^{4}=\ ^{n}C_{5}(a)^{n-5}(b)^{5}$

$\frac{a}{b}=\frac{^{n}C_{5}}{^{n}C_{4}} =\frac{n-5+1}{5}$

$\frac{a}{b}=\frac{n-4}{5}$

Correct option is 2.

Option 1)

$\frac{n-5}{6}\;$

This is an incorrect option.

Option 2)

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

This is the correct option.

Option 3)

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

This is an incorrect option.

Option 4)

$\; \frac{6}{n-5}$

This is an incorrect option.

Posted by

divya.saini

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