Formula for Binomial Theorem

Here you will learn formula for binomial theorem of class 11 with examples.

Let’s begin –

Formula for Binomial Theorem

If x and a are real numbers, then for all n \(\in\) N.

\((x + a)^n\) = \(^{n}C_0 x^n a^0\) + \(^{n}C_1 x^{n – 1} a^1\) + …………… + \(^{n}C_r x^{n – r} a^r\) + …………… + \(^{n}C_{n – 1} x^1 a^{n – 1}\) + \(^{n}C_n x^0 a^n\)

i.e. \((x + a)^n\) = \(\sum_{r=0}^{n}\) \(^{n}C_{r} x^{n-r} a^{r}\)

Some Importants Result From Binomial Theorem

(1) We have.

\((x + a)^n\) = \(\sum_{r=0}^{n}\) \(^{n}C_{r} x^{n-r} a^{r}\).

or, \((x + a)^n\) = \(^{n}C_0 x^n a^0\) + \(^{n}C_1 x^{n – 1} a^1\) + …………… + \(^{n}C_r x^{n – r} a^r\) + …………… + \(^{n}C_n x^0 a^n\)

Since r can have values from 0 to n, the total number of terms in the expansion is (n + 1).

(2) The sum of the indices of x and a in each term is n.

(3) Since \(^{n}C_r\) = \(^{n}C_{n – r}\)

\(\implies\) \(^{n}C_0\) = \(^{n}C_n\), \(^{n}C_1\) = \(^{n}C_{n-1}\), \(^{n}C_2\) = \(^{n}C_{n – 2}\) = …….

So, the coefficients of terms equidistant from the beginning and end are equal. These coefficients are known as binomial coefficients.

(4) Replacing a by -a, we get

\((x – a)^n\) = \(^{n}C_0 x^n a^0\) – \(^{n}C_1 x^{n – 1} a^1\) + …………… + \((-1)^r\)\(^{n}C_r x^{n – r} a^r\) + …………… + \((-1)^n\)\(^{n}C_n x^0 a^n\)

i.e.  \((x – a)^n\) = \(\sum_{r=0}^{n}\) \((-1)^r\) \(^{n}C_{r} x^{n-r} a^{r}\)

Thus, the terms in the expansion of \((x – a)^n\) are alternatively positive and negative, the last term is positive or negative according as n is even or odd.

(5) Putting x = 1 and a = x in the expansion of \((x + a)^n\), we get

\((1 + x)^n\) = \(^{n}C_0\) + \(^{n}C_1 x\) + \(^{n}C_2 x^2\) + ……… + \(^{n}C_r x^r\) + \(^{n}C_n x^n\)

i.e.  \((1 + x)^n\) = \(\sum_{r=0}^{n}\) \(^{n}C_r x^r\)

This is the expansion of \((1 + x)^n\) in ascending powers of x.

(6) Putting a = 1 in the expansion of \((x + a)^n\), we get.

\((1 + x)^n\) = \(^{n}C_0 x^n\) + \(^{n}C_1 x^{n-1}\) + \(^{n}C_2 x^{n-2}\) + ……… + \(^{n}C_r x^{n-r}\) + \(^{n}C_{n-1} x\) + \(^{n}C_n\).

i.e.  \((1 + x)^n\) = \(\sum_{r=0}^{n}\) \(^{n}C_r x^{n-r}\)

This is the expansion of \((1 + x)^n\) in descending powers of x.

(7) Putting x = 1 and a = -x in the expansion of \((x + a)^n\), we get

\((1 – x)^n\) = \(^{n}C_0\) – \(^{n}C_1 x\) + \(^{n}C_2 x^2\) + ……… + \((-1)^r\)\(^{n}C_r x^r\) + \((-1)^n\) \(^{n}C_n x^n\).

i.e.  \((1 – x)^n\) = \(\sum_{r=0}^{n}\) \((-1)^n\) \(^{n}C_r x^r\)

(8) The coefficient of (r + 1)th term in the expansion of \((1 + x)^n\) is \(^{n}C_r\).

(9) The coefficient of \(x^r\) in the expansion of \((1 + x)^n\) is \(^{n}C_r\).

Example : Expand \((x^2 + 2a)^5\) by binomial theorem.

Solution : Using binomial theorem, we have

\((x^2 + 2a)^5\) = \(^5C_0 (x^2)^5 (2a)^0\) + \(^5C_1 (x^2)^4 (2a)^1\) + \(^5C_2 (x^2)^3 (2a)^2\) + \(^5C_3 (x^2)^2 (2a)^3\) + \(^5C_4 (x^2) (2a)^4\) + \(^5C_5 (x^2)^0 (2a)^5\)

= \(x^{10}\) + 5\((x^8)(2a)\) + 10\((x^6)(4a^2)\) + 10\((x^4)(8a^3)\) + 5\((x^2)(16a^4)\) + 325\((a^5)\)

=  \(x^{10}\) + 10\(x^{8}a\) + 40\(x^{6}a^2\) + 80\(x^{4}a^3\) + 80\(x^{2}a^4\) + 32\(a^{5}\)


Related Questions

By using binomial theorem, expand \((1 + x + x^2)^3\).

Which is larger \((1.01)^{1000000}\) or 10,000?

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