Using binominal theorem, evaluate each of the following :
(i) $(101)^{4}$
(ii) $(98)^{4}$
(iii) $(1.2)^{4}$
(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
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