If a and b are coefficients of xⁿ in the expansions of ${(1 + x)}^{2 n} and {(1 + x)}^{2 n - 1}$ respectively, then write the relation between a and b.

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Solution

$Coefficient of x^{n} in the expansion (1 + x)^{2 n} = {}^{2 n}C_{n} = a Coefficient of x^{n} in the expansion (1 + x)^{2 n - 1} = {}^{2 n - 1}C_{n} = b Now, we have : {}^{2 n}C_{n} = \frac{2 n!}{n! . n!} = \frac{2 n (2 n - 1)!}{n (n - 1)! n!} . . . (1) and {}^{2 n - 1}C_{n} = \frac{(2 n - 1)!}{n! (n - 1)!} . . . (2) Dividing equation (1) by (2), we get \Rightarrow \frac{{}^{2 n}C_{n}}{{}^{2 n - 1}C_{n}} = \frac{2 n (2 n - 1)! n! (n - 1)!}{n (n - 1)! n! (2 n - 1)!} \Rightarrow \frac{a}{b} = 2 \Rightarrow a = 2 b$

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If a and b are coefficients of xn in the expansions of 1+x2n and 1+x2n-1 respectively, then write the relation between a and b.

Coefficient of xn in the expansion (1+x)2n = Cn2n=aCoefficient of xn in the expansion (1+x)2n-1=Cn2n-1=bNow, we have: Cn2n=2n!n!. n!=2n(2n-1)!nn-1! n! ...1 and Cn2n-1=(2n-1)!n!(n-1)! ...2Dividing equation 1 by 2, we get⇒ Cn2nCn2n-1=2n(2n-1)! n! (n-1)!nn-1! n! (2n-1)!⇒ab=2⇒a=2b

If a and b are coefficients of xⁿ in the expansions of ${(1 + x)}^{2 n} and {(1 + x)}^{2 n - 1}$ respectively, then write the relation between a and b.