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

Binomial Theorem PYQ — JEE Main 2007 · Shift 0 · Q59 with Solution

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Correct answer: B

For

(a-b)^n= \sum_{r=0}^{n} \binom{n}{r} a^{n-r}(-b)^r,

the $(r+1)^{\text{th}}$ term is

T_{r+1}=\binom{n}{r}a^{n-r}(-b)^r.

T_5=\binom{n}{4}a^{n-4}b^4

since $(-b)^4=b^4$ .

T_6=\binom{n}{5}a^{n-5}(-b)^5=-\binom{n}{5}a^{n-5}b^5.

Given

T_5+T_6=0,

we get

\binom{n}{4}a^{n-4}b^4-\binom{n}{5}a^{n-5}b^5=0.

Factor out $a^{n-5}b^4$ :

a^{n-5}b^4\left[\binom{n}{4}a-\binom{n}{5}b\right]=0.

Assuming $a,b\neq 0$ , this gives

\binom{n}{4}a=\binom{n}{5}b.

Hence

\frac{a}{b}=\frac{\binom{n}{5}}{\binom{n}{4}}.

\frac{\binom{n}{5}}{\binom{n}{4}} = \frac{\dfrac{n!}{5!(n-5)!}}{\dfrac{n!}{4!(n-4)!}} = \frac{4!(n-4)!}{5!(n-5)!} = \frac{n-4}{5}.

So,

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

This is Option B.

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

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View written solutionFree

Correct answer: B

For

(a-b)^n= \sum_{r=0}^{n} \binom{n}{r} a^{n-r}(-b)^r,

the $(r+1)^{\text{th}}$ term is

T_{r+1}=\binom{n}{r}a^{n-r}(-b)^r.

T_5=\binom{n}{4}a^{n-4}b^4

since $(-b)^4=b^4$ .

T_6=\binom{n}{5}a^{n-5}(-b)^5=-\binom{n}{5}a^{n-5}b^5.

Given

T_5+T_6=0,

we get

\binom{n}{4}a^{n-4}b^4-\binom{n}{5}a^{n-5}b^5=0.

Factor out $a^{n-5}b^4$ :

a^{n-5}b^4\left[\binom{n}{4}a-\binom{n}{5}b\right]=0.

Assuming $a,b\neq 0$ , this gives

\binom{n}{4}a=\binom{n}{5}b.

Hence

\frac{a}{b}=\frac{\binom{n}{5}}{\binom{n}{4}}.

\frac{\binom{n}{5}}{\binom{n}{4}} = \frac{\dfrac{n!}{5!(n-5)!}}{\dfrac{n!}{4!(n-4)!}} = \frac{4!(n-4)!}{5!(n-5)!} = \frac{n-4}{5}.

So,

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

This is Option B.

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