mathematics-class-11-unit-09-chapter-01-binomial-theorem-l-3-9-kszvuj-qeba

### <Success>Binomial theorem</Success>
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<p>$ (a+b)^n ={}^nC_0 a^n+{ }^n C_1 a^{n-1} b+{ }^n C_2 a^{n-2} b^2 $</p>
<p>$ +{ }^n C_3 a^{n-3} b^3+\cdots \cdot{ }^n C_{n-1} a b^{n-1}+ { }^n C_n b^n $</p>
<p>$\quad =\sum_{k=0}^n{ }^n C_k a^{n-k} b^k $</p>
<p><strong>Example</strong> $ (a-b)^7$</p>
<p>$ (a-b)^7= a^7-7 a^6 b+21 a^5 b^2-35 a^4 b^3 +35 a^3 b^4-21 a^2 b^5+7 a b^6-b^7 $</p>
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<h3 id="successapplication-interest-problemsuccess"><Success>Application: interest problem</Success></h3>
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<p>$ 6 $ % p.a.$ \quad P \rightarrow 1.06 P \rightarrow 1.06 \times 1.06 P $</p>
<p>$ (1.06)^{20} \mathrm{P} $</p>
<p><strong>Solution:</strong>   $ (1+0.06)^{20}=1 + 20 \times 0.06+{ }^{20} C_2 0.06^2 +{ }^{20} C_3 0.06^3+{ }^{20} C_4 0.06^4+\ldots $</p>
<p>$\biggl [ { }^{20} C_2=\frac{20 \times 19}{2}=190 $</p>
<p>$ { }^{20} C_3=\frac{20 \times 19 \times 18}{2 \times 3}=190 \times 6=1140 $</p>
<p>$ { }^{20} C_4={ }^{20} C_3 \times \frac{17}{4} $</p>
<p>$ { }^{20} C_5={ }^{20} C_4 \times \frac{16}{5}\biggl] $</p>
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