What went wrong when the king rode under the arch?

It was too low and knocked off his crown

How does the chain of blame unfold in the poem?

Each accused person blames another, ending with the king's own changes

State the binomial theorem for a positive integer \(n\).

\((a+b)^n={}^{n}C_0a^n+{}^{n}C_1a^{n-1}b+\cdots+{}^{n}C_n b^n=\sum_{k=0}^{n}{}^{n}C_k a^{n-k}b^k\).

How many terms are in the expansion of \((a+b)^n\)?

There are \(n+1\) terms, one more than the index \(n\).

What is Pascal's triangle?

A triangular arrangement of binomial coefficients in which 1s border each row and every interior entry equals the sum of the two entries above it.

\(x^6+12x^5+60x^4+160x^3+240x^2+192x+64\).

Show that \(\sum_{k=0}^{n}{}^{n}C_k=2^n\).

Put \(a=b=1\) in the theorem (or \(x=1\) in \((1+x)^n\)): \((1+1)^n=\sum {}^{n}C_k=2^n\).

Write the expansion of \((x-y)^n\).

\((x-y)^n={}^{n}C_0x^n-{}^{n}C_1x^{n-1}y+{}^{n}C_2x^{n-2}y^2-\cdots+(-1)^n{}^{n}C_n y^n\).

Evaluate \((98)^5\) using the theorem.

Write \(98=100-2\) and expand \((100-2)^5\) to get \(9039207968\).

State the property of binomial coefficients used to build Pascal's triangle.

\({}^{n}C_r+{}^{n}C_{r-1}={}^{n+1}C_r\), so each row is generated from the previous one.

Put \(x=-1\) in \((1+x)^n\): what identity results?

\(0={}^{n}C_0-{}^{n}C_1+{}^{n}C_2-\cdots+(-1)^n{}^{n}C_n\), the alternating sum of binomial coefficients is zero.

Why is \((1.01)^{1000000}>10000\)?

Writing \((1+0.01)^{1000000}=1+1000000(0.01)+\text{positive terms}=1+10000+\cdots>10000\).

Binomial Theorem

Class 11 Mathematics · Chapter 7

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Every formula of Ch 1–14, with its condition

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25 marks: Unit II, Ch 4–8
23 marks: Unit I, Ch 1–3
Common traps, corrected
Conic table and standard values
Graphs plotted from the book's own tables

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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.

Binomial theorem for positive integral indicesBinomial coefficientsPascal's triangleGeneral observations on the expansionSpecial cases \((1\pm x)^n\)Applications: evaluation and divisibility

Key terms

Binomial theorem

\((a+b)^n=\sum_{k=0}^{n}{}^{n}C_k a^{n-k}b^k\) for positive integer \(n\).

Binomial coefficient

The coefficient \({}^{n}C_k=\dfrac{n!}{k!\,(n-k)!}\) of each term.

Pascal's triangle

A triangular array of binomial coefficients; each entry is the sum of the two directly above it.

Number of terms

The expansion of \((a+b)^n\) has \(n+1\) terms.

Index sum

In each term the powers of \(a\) and \(b\) add up to \(n\).

Special case

\((1+x)^n={}^{n}C_0+{}^{n}C_1 x+\cdots+{}^{n}C_n x^n\).

Extra questions & answers

Try one from this chapter

The number of terms in the expansion of \((a+b)^n\) is:

Flashcards & interactive practice