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Prove that C_{1}+2 C_{2}+3 C_{3}+4 C_{4}+.+ ^nC_{n}=n.2^{n-1}

Question

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

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Prove that

C_{1} + 2 C_{2} + 3 C_{3} + 4 C_{4} + . +^{n} C_{n} = n . 2^{n - 1}

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Text solutionVerified

C_{1} + 2 C_{2} + 3 C_{3} + .... + n C_{n}

= n + \frac{2 n ( n - 1 )}{2 !} + \frac{3 n ( n - 1 ) ( n - 2 )}{3 !} + ..... + n \times 1

= n + n (n - 1) + \frac{n ( n - 1 ) ( n - 2 )}{2} + ...... + n

= n [1 + n - 1 + \frac{( n - 1 ) ( n - 2 )}{2} + ..... + 1]

= n [1 + \frac{n - 1}{1 !} + \frac{( n - 1 ) ( n - 2 )}{2} + ..... + 1]

put

n - 1 = N

= n [1 + \frac{N}{1 !} + \frac{N ( N + 1 )}{2 !} + ..... + 1]

= n [N C_{0} + N C_{1} + N C_{2} + ...... + N C_{N}]

= n 2^{N}

= n 2^{n - 1}

Hence proved.

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