In the binomial expansion of (a-b)^{n},n\geq 5, then sum of 5th and 6th 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)

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.

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