The coefficient of x4 in the binomial expansion of 1-x4-1-x5-1-x11-1-bx12 is n-1-13C5 for some n - Then the smallest natural number possible for b is

(1+x)^4+(1+x)^5+⋯+(1+x)^11+(1+bx)^12 =(1+x)^4 [(1+x)^8 1](1+x) 1+(1+bx)^12 =(1+x)^12x (1+x)^4x+(1+bx)^12 Coefficient of x^4=^12C_5 0+^12C_4b^4 So, ^12C_5+^12C_ ...

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Standard XII

Mathematics

General Term of Binomial Expansion

The coefficie...

Question

The coefficient of $x^{4}$ in the binomial expansion of $(1 + x)^{4} + (1 + x)^{5} + \dots + (1 + x)^{11} + (1 + b x)^{12}$ is $(n + 1) \cdot^{13} C_{5}$ for some $n \in N$ . Then the smallest natural number possible for $b$ is

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Solution

$(1 + x)^{4} + (1 + x)^{5} + \dots + (1 + x)^{11} + (1 + b x)^{12}$
$= \frac{(1 + x)^{4} [(1 + x)^{8} - 1]}{(1 + x) - 1} + (1 + b x)^{12}$
$= \frac{(1 + x)^{12}}{x} - \frac{(1 + x)^{4}}{x} + (1 + b x)^{12}$

Coefficient of $x^{4} =^{12} C_{5} - 0 +^{12} C_{4} b^{4}$
So, $^{12} C_{5} +^{12} C_{4} b^{4} = (n + 1) \cdot^{13} C_{5}$
$\Rightarrow b^{4} - 1 = \frac{13}{5} n$
$\Rightarrow n = \frac{5 (b^{4} - 1)}{13}$
$\Rightarrow n = \frac{5 (b - 1) (b + 1) (b^{2} + 1)}{13}$
Since $n \in N$ , the least possible value of $b$ is $12$

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The coefficient of x4 in the binomial expansion of (1+x)4+(1+x)5+⋯+(1+x)11+(1+bx)12 is (n+1)⋅13C5 for some n∈N. Then the smallest natural number possible for b is

(1+x)4+(1+x)5+⋯+(1+x)11+(1+bx)12 =(1+x)4[(1+x)8−1](1+x)−1+(1+bx)12 =(1+x)12x−(1+x)4x+(1+bx)12 Coefficient of x4=12C5−0+12C4b4 So, 12C5+12C4b4=(n+1)⋅13C5 ⇒b4−1=135n ⇒n=5(b4−1)13 ⇒n=5(b−1)(b+1)(b2+1)13 Since n∈N, the least possible value of b is 12

The coefficient of $x^{4}$ in the binomial expansion of $(1 + x)^{4} + (1 + x)^{5} + \dots + (1 + x)^{11} + (1 + b x)^{12}$ is $(n + 1) \cdot^{13} C_{5}$ for some $n \in N$ . Then the smallest natural number possible for $b$ is