If the number of terms in the expansion of ( 1 - (2/x) + (4/x2) )n , x ≠ 0 , is 28 , then the sum of the coefficients of all the terms in this expansion, is :

Question

If the number of terms in the expansion of ( 1 - (2/x) + (4/x2) )n , x ≠ 0 , is 28 , then the sum of the coefficients of all the terms in this expansion, is : (A) 64 (B) 2187 (C) 243 (D) 729

Tardigrade Admin · Accepted Answer

Number of terms = ((n +1)(n+2)/2) = 28 ⇒ n = 6 ∴ a0 + (a1/x) + (a2/x2) + .... + (a2n/x2n) = ( 1 - (2/x) + (4/x2) )n Put x = 1, n = 6 , a0 + a1 + a2 + ... + a2n = 36 = 729

Q. If the number of terms in the expansion of $(1 - \frac{2}{x} + \frac{4}{x ^{2}})^{n}, x \neq = 0$ , is $28$ , then the sum of the coefficients of all the terms in this expansion, is :

64

2187

243

729

Number of terms $= \frac{( n + 1 ) ( n + 2 )}{2} = 28$
$\Rightarrow n = 6$
$∴ a_{0} + \frac{a _{1}}{x} + \frac{a _{2}}{x ^{2}} + .... + \frac{a _{2} n}{x ^{2 n}} = (1 - \frac{2}{x} + \frac{4}{x ^{2}})^{n}$
Put $x = 1, n = 6, a_{0} + a_{1} + a_{2} + ... + a_{2 n} = 3^{6} = 729$

Q. If the number of terms in the expansion of (1−x2​+x24​)n,x=0 , is 28 , then the sum of the coefficients of all the terms in this expansion, is :

64

2187

243

729

Number of terms =2(n+1)(n+2)​=28 ⇒n=6 ∴a0​+xa1​​+x2a2​​+....+x2na2​n​=(1−x2​+x24​)n Put x=1,n=6,a0​+a1​+a2​+...+a2n​=36=729

Q. If the number of terms in the expansion of $(1 - \frac{2}{x} + \frac{4}{x ^{2}})^{n}, x \neq = 0$ , is $28$ , then the sum of the coefficients of all the terms in this expansion, is :

Number of terms $= \frac{( n + 1 ) ( n + 2 )}{2} = 28$
$\Rightarrow n = 6$
$∴ a_{0} + \frac{a _{1}}{x} + \frac{a _{2}}{x ^{2}} + .... + \frac{a _{2} n}{x ^{2 n}} = (1 - \frac{2}{x} + \frac{4}{x ^{2}})^{n}$
Put $x = 1, n = 6, a_{0} + a_{1} + a_{2} + ... + a_{2 n} = 3^{6} = 729$