4C1+5C2(12)+6C3(12)2+−−−−∞=

Let S=4C1+5C2(12)+6C3(12)2+...... S2=4(12)+10(12)2+20(12)3+..... 1+S2=1+4(12)+10(12)2+20(12)3+..... 1+S2=(1−12)−4 ∴S=30

4C1+5C2(12)+6C3(12)2+−−−−∞=

Let S=4C1+5C2(12)+6C3(12)2+...... S2=4(12)+10(12)2+20(12)3+..... 1+S2=1+4(12)+10(12)2+20(12)3+..... 1+S2=(1−12)−4 ∴S=30

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$4_{C_{1}} + 5_{C_{2}} (\frac{1}{2}) + 6_{C_{3}} {(\frac{1}{2})}^{2} + - - - - \infty =$

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answer is C.

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Detailed Solution

Let $S = 4_{C_{1}} + 5_{C_{2}} (\frac{1}{2}) + 6_{C_{3}} {(\frac{1}{2})}^{2} + ......$
$\frac{S}{2} = 4 (\frac{1}{2}) + 10 {(\frac{1}{2})}^{2} + 20 {(\frac{1}{2})}^{3} + ..... 1 + \frac{S}{2} = 1 + 4 (\frac{1}{2}) + 10 {(\frac{1}{2})}^{2} + 20 {(\frac{1}{2})}^{3} + ..... 1 + \frac{S}{2} = {(1 - \frac{1}{2})}^{- 4} ∴ S = 30$