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The Binomial Expansion

The Binomial Expansion

The binomial series associates a finite series with an expression of the for $(a+b)^n$, where $n$ is a natural number The formula for this series is $$ (a+b)^n = \sum_{r=0}^n {}^n\!C_ra^rb^{n-r} $$

The Binomial Expansion 1

Determine the expansion of the following expression $(3-5t)^4$
solution - press button to display
$$ \begin{array}{rcl} (3-5t)^4 &=& {}^{4}C_0(3)^0(-5t)^4 + {}^{4}C_1(3)^1(-5t)^3 + {}^{4}C_2(3)^2(-5t)^2 \\ & &+{}^{4}C_3(3)^3(-5t)^1+ {}^{4}C_4(3)^4(-5t)^0 \\ &=&625t^4 - 1500t^3 + 1350t^2 - 540t + 81 \end{array} $$

The Binomial Expansion 2

Determine the coefficient of $t^{11}$ in the expansion of $$ \left(t^2 + \frac{2}{t}\right)^{10} $$
solution - press button to display
The general term in the expansion will be of the form $$ \begin{array}{rcl} ^{10}C_k (t^2)^k\left(\frac{2}{t}\right)^{10-k} &=& ^{10}C_k (t^{2k})2^{10-k}t^{k-10} \\ &=& ^{10}C_k 2^{10-k} t^{3k-10} \end{array} $$ Given that we want the coefficient of $t^{11}$, then we require that $3k-10 = 11$ which implies that $k=7$. The coefficient of $t^{11}$ is therefore $ ^{10}C_7 \times 2^3 = 960$

The Binomial Expansion 3

Given that, in the expansion of $(1-t)(4+t)^n$ the coefficient of $t$ is $256$, determine the value of $n$
solution - press button to display
The coefficient of $t$ is comprised of two parts, we get the coefficient arising from $-t\times 4^n$ and the coefficient arising from $1\times 4^{n-1}\times nt$. The coefficient is therefore $$ -4^n + 4^{n-1}n = 256 $$ A little work shows this equation admits a unique solution of $n=5$.