Chapter 7: Binomial TheoremClass 11 Mathematics — summary, notes, extra questions & MCQ quiz

Summary

The binomial theorem gives an efficient way to expand \((a+b)^n\) for a positive integer \(n\) without repeated multiplication. The expansion is \((a+b)^n={}^{n}C_0a^n+{}^{n}C_1a^{n-1}b+{}^{n}C_2a^{n-2}b^2+\cdots+{}^{n}C_n b^n\), written compactly as \(\sum_{k=0}^{n}{}^{n}C_k a^{n-k}b^k\). It has \(n+1\) terms; the power of \(a\) decreases from \(n\) to \(0\) while the power of \(b\) increases from \(0\) to \(n\), and the index sum stays \(n\) in every term. The coefficients \({}^{n}C_k\) are the binomial coefficients, the same numbers that form Pascal's triangle, where each entry is the sum of the two above it. Special cases include \((1+x)^n=\sum {}^{n}C_k x^k\), so that putting \(x=1\) gives \(\sum {}^{n}C_k=2^n\) and \(x=-1\) gives the alternating sum \(0\). The binomial coefficients are generated by the Pascal rule \({}^{n}C_r+{}^{n}C_{r-1}={}^{n+1}C_r\), so each row of the triangle can be written from the one above without expanding. The theorem is proved by the principle of mathematical induction. It is used to compute powers like \((98)^5\) and \((101)^6\), compare magnitudes such as \((1.01)^{1000000}\) versus \(10000\), and prove divisibility results, for instance that \(6^n-5n\) leaves remainder 1 when divided by 25.

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