when (1+bx)^5 is expanded the coefficient of x is equal to the coefficient of x^2. Find the value of b when b is not equal to zero

Question

When $$(1+bx)^5$$ is expanded the coefficient of $x$ is equal to the coefficient of $x^2$. Find the value of $b$ when $b$ is not equal to zero

I tried to solve it by using the $T_{r+1}$ formula as well as the binomial theorem formula. But I just can't understand this. Anyone help

Answer 1

You can easily notice that $$[x]=\binom51b^1=5b$$ and $$\left[x^2\right]=\binom52b^2=10b^2$$

Therefore, $$5b=10b^2$$ Since $b\ne0$, $$5=10b$$ giving $$b=\frac12$$

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For verification,

$$(1+0.5x)^5=0.03125 x^5 + 0.3125 x^4 + 1.25 x^3 + \color{red}{2.5} x^2 + \color{red}{2.5} x + 1$$

Answer 2

According to the Taylor polynomial expansion of $$ f(x)= (1+bx)^5$$ we get

$$ f(x)= f(0)+f'(0)x+f"(0) x^2/{2!}+...+f^{(5)}(0) x^5/{5!}$$

$$ f(x)= (1+bx)^5 \implies \\f'(0)=5b, f''(0)/2=10 b^2$$

$$ 5b=10b^2 \implies b=1/2$$