# 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?