Using binominal theorem, evaluate each of the following :

Solution:

(i) $(101)^{4}$

To find: Value of $(101)^{4}$

Formula used: (i) ${ }^{n} C_{r}=\frac{n !}{(n-r) !(r) !}$

(ii) $(a+b)^{n}={ }^{n} C_{0} a^{n}+{ }^{n} C_{1} a^{n-1} b+{ }^{n} C_{2} a^{n-2} b^{2}+\ldots \ldots+{ }^{n} C_{n-1} a b^{n-1}+{ }^{n} C_{n} b^{n}$

$101=(100+1)$

Now $(101)^{4}=(100+1)^{4}$

$(100+1)^{4}=$

$\left[{ }^{4} C_{0}(100)^{4-0}\right]+\left[{ }^{4} C_{1}(100)^{4-1}(1)^{1}\right]+\left[{ }^{4} C_{2}(100)^{4-2}(1)^{2}\right]+$

$\left[{ }^{4} C_{3}(100)^{4-3}(1)^{3}\right]+\left[{ }^{4} C_{4}(1)^{4}\right]$

$\Rightarrow\left[{ }^{4} C_{0}(100)^{4}\right]+\left[{ }^{4} C_{1}(100)^{3}(1)^{1}\right]+\left[{ }^{4} C_{2}(100)^{2}(1)^{2}\right]+$

$\left[{ }^{4} C_{3}(100)^{1}(1)^{3}\right]+\left[{ }^{4} C_{4}(1)^{4}\right]$

$\Rightarrow\left[\frac{4 !}{0 !(4-0) !}(100000000)\right]+\left[\frac{4 !}{1 !(4-1) !}(1000000)\right]+$

$\left[\frac{4 !}{2 !(4-2) !}(10000)\right]+\left[\frac{4 !}{3 !(4-3) !}(100)\right]+\left[\frac{4 !}{4 !(4-4) !}(1)\right]$

$\Rightarrow[(1)(100000000)]+[(4)(1000000)]+[(6)(10000)]+$

$[(4)(100)]+[(1)(1)]$

= 104060401

Ans) 104060401

(ii) $(98)^{4}$

To find: Value of $(98)^{4}$

Formula used: (I) ${ }^{n} C_{r}=\frac{n !}{(n-r) !(r) !}$

(ii) $(a+b)^{n}={ }^{n} C_{0} a^{n}+{ }^{n} C_{1} a^{n-1} b+{ }^{n} C_{2} a^{n-2} b^{2}+\ldots \ldots+{ }^{n} C_{n-1} a b^{n-1}+{ }^{n} C_{n} b^{n}$

$98=(100-2)$

Now $(98)^{4}=(100-2)^{4}$

$(100-2)^{4}$

$=\left[{ }^{4} C_{0}(100)^{4-0}\right]+\left[{ }^{4} C_{1}(100)^{4-1}(-2)^{1}\right]+\left[{ }^{4} C_{2}(100)^{4-2}(-2)^{2}\right]+$

$\left[{ }^{4} C_{3}(100)^{4-3}(-2)^{3}\right]+\left[{ }^{4} C_{4}(-2)^{4}\right]$

$\Rightarrow\left[{ }^{4} C_{0}(100)^{4}\right]-\left[{ }^{4} C_{1}(100)^{3}(2)\right]+\left[{ }^{4} C_{2}(100)^{2}(4)\right]-\left[{ }^{4} C_{3}(100)^{1}(8)\right]+$

$\left[{ }^{4} C_{4}(16)\right]$

$\Rightarrow\left[\frac{4 !}{0 !(4-0) !}(100000000)\right]-\left[\frac{4 !}{1 !(4-1) !}(1000000)(2)\right]+$

$\left[\frac{4 !}{2 !(4-2) !}(10000)(4)\right]-\left[\frac{4 !}{3 !(4-3) !}(100)(8)\right]+\left[\frac{4 !}{4 !(4-4) !}(16)\right]$

$\Rightarrow[(1)(100000000)]-[(4)(1000000)(2)]+[(6)(10000)(4)]-$

$[(4)(100)(8)]+[(1)(16)]$

= 92236816

Ans) 92236816

(iii) $(1.2)^{4}$

To find: Value of $(1.2)^{4}$

Formula used: (i) ${ }^{n} C_{r}=\frac{n !}{(n-r) !(r) !}$

(ii) $(a+b)^{n}={ }^{n} C_{0} a^{n}+{ }^{n} C_{1} a^{n-1} b+{ }^{n} C_{2} a^{n-2} b^{2}+\ldots \ldots+{ }^{n} C_{n-1} a b^{n-1}+{ }^{n} C_{n} b^{n}$

$1.2=(1+0.2)$

Now $(1.2)^{4}=(1+0.2)^{4}$

$(1+0.2)^{4}$

$=\left[{ }^{4} C_{0}(1)^{4-0}\right]+\left[{ }^{4} C_{1}(1)^{4-1}(0.2)^{1}\right]+\left[{ }^{4} C_{2}(1)^{4-2}(0.2)^{2}\right]+$

$\left.{ }^{4} C_{3}(1)^{4-3}(0.2)^{3}\right]+\left[{ }^{4} C_{4}(0.2)^{4}\right]$

$\Rightarrow\left[{ }^{4} C_{0}(1)^{4}\right]+\left[{ }^{4} C_{1}(1)^{3}(0.2)^{1}\right]+\left[{ }^{4} C_{2}(1)^{2}(0.2)^{2}\right]+\left[{ }^{4} C_{3}(1)^{1}(0.2)^{3}\right]+$

$\left[{ }^{4} C_{4}(0.2)^{4}\right]$

$\Rightarrow\left[\frac{4 !}{0 !(4-0) !}(1)\right]+\left[\frac{4 !}{1 !(4-1) !}(1)(0.2)\right]+\left[\frac{4 !}{2 !(4-2) !}(1)(0.04)\right]+$

$\left[\frac{4 !}{3 !(4-3) !}(1)(0.008)\right]+\left[\frac{4 !}{4 !(4-4) !}(0.0016)\right]$

$\Rightarrow[(1)(1)]+[(4)(1)(0.2)]+[(6)(1)(0.04)]+[(4)(1)(0.008)]+$

$[(1)(0.0016)]$

= 2.0736

Ans) 2.0736