Prove that $C_{0} + 2. C_{1} + 4. C_{2} + 8. C_{3} + . . . . . . + 2^{n} . C_{n} = 3^{n}$ .

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Solution

Fact: $(1 + x)^{n} =^{n} C_{0} +^{n} C_{1} x +^{n} C_{2} x^{2} +^{n} C_{3} x^{3} + . . . . . . . . . . . . +^{n} C_{n} x^{n}$
Substitute $x = 2$ in above identity
$(1 + 2)^{n} =^{n} C_{0} +^{n} C_{1} (2) +^{n} C_{2} (2)^{2} +^{n} C_{3} (2)^{3} + . . . . . . . . . . . . +^{n} C_{n} (2)^{n}$
$∴^{n} C_{0} + 2 \cdot^{n} C_{1} + 4 \cdot^{n} C_{2} + 8 \cdot^{n} C_{3} + . . . . . . . . + 2^{n} \cdot^{n} C_{n} = 3^{n}$
Hence proved

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