Question

The value of (402)^½ according to binomial theorem is
$\sqrt{402}$
​

Answer 1

Given :    $(402)^{\frac{1}{2} }$

To Find : The value  according to the binomial theorem

20.05

20.01

20.00

20.2​

Solution:

(1 + x)ⁿ  = 1  +  nx  + n(n-1)x² /2!  + ...

where n  is fraction p/q form   and x  <  1

$(402)^{\frac{1}{2} }$

$=(400+2)^{\frac{1}{2} }$

$=(400)^{\frac{1}{2} }(1+\frac{2}{400} )^{\frac{1}{2} }$

∵ √400  = 20

$=20(1+\frac{1}{200} )^{\frac{1}{2} }$

x = 1/200  and  n = 1/2

as x is very small so term like x² and further can be ignored

so (1 + x)ⁿ  ≈  1  +  nx

= 20 ( 1  +  (1/2)(1/200))

= 20  + 1/20

= 20 + 0.05

= 20.05

Value of  $(402)^{\frac{1}{2} }$  is about 20.05

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