Question
Expand $(1+x)^{6}$ and verify it by putting $x=1$ on both the sides.

Answer

$(1+x)^{6}$
$={ }^{6} \mathrm{C}_{0}(1)^{6}(x)^{0}+{ }^{6} \mathrm{C}_{1}(1)^{5}(x)^{1}+{ }^{6} \mathrm{C}_{2}(1)^{4}(x)^{2}+{ }^{6} \mathrm{C}_{3}(1)^{3}(x)^{3} $
$ +{ }^{6} \mathrm{C}_{4}(1)^{2}(x)^{4}+{ }^{6} \mathrm{C}_{5}(1)^{1}(x)^{5}+{ }^{6} \mathrm{C}_{6}(1)^{0}(x)^{6} $
$=1+6 x+15 x^{2}+20 x^{3}+15 x^{4}+6 x^{5}+x^{6}$
$\text { Thus, } $
$(1+x)^{6}=1+6 x+15 x^{2}+20 x^{3}+15 x^{4}+6 x^{5}+x^{6} \\ \text { Putting } x=1$
$ \text { L.H.S. }=(1+x)^{6}=(1+1)^{6}=(2)^{6}=64 $
$ \text { R.H.S. }=1+6 x+15 x^{2}+20 x^{3}+15 x^{4}+6 x^{5}+x^{6}$
$=1+6(1)+15(1)^{2}+20(1)^{3}+15(1)^{4}+6(1)^{5}+(1)^{6} $
$ =1+6+15+20+15+6+1$
$ =64$
$ \text { Thus, L.H.S. }=\text { R.H.S. }$

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