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Standard XII

Mathematics

Multinomial Expansion

If a and b ar...

Question

If a and b are the coefficients of xⁿ in the expansion of ${(1 + x)}^{2 n} and {(1 + x)}^{2 n - 1}$ respectively, find $\frac{a}{b}$ .

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Solution

Coefficients of xⁿ in the expansion of ${(1 + x)}^{2 n}$ is $^{2 n} C_{n} = a$ .

Coefficients of xⁿ in the expansion of ${(1 + x)}^{2 n - 1}$ is $^{2 n - 1} C_{n} = b$ .

Now,
$\frac{a}{b} = \frac{^{2 n} C_{n}}{^{2 n - 1} C_{n}} = \frac{\frac{(2 n)!}{n! n!}}{\frac{(2 n - 1)!}{n! (n - 1)!}} = \frac{2 n}{n} = 2$

Hence, $\frac{a}{b} = 2$ .

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If a and b are the coefficients of xn in the expansion of 1+x2n and 1+x2n-1 respectively, find ab.

Coefficients of xn in the expansion of 1+x2n is 2nCn=a. Coefficients of xn in the expansion of 1+x2n-1 is 2n-1Cn=b. Now, ab=2nCn2n-1Cn =2n!n!n!2n-1!n!n-1! =2nn =2 Hence, ab=2.

If a and b are the coefficients of xⁿ in the expansion of ${(1 + x)}^{2 n} and {(1 + x)}^{2 n - 1}$ respectively, find $\frac{a}{b}$ .