If {C_0}, {C_1}, {C_2}, ......., {C_n} are th | vedclass

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(B) We know the binomial expansions:$(1 + x)^n = {C_0} + {C_1}x + {C_2}x^2 + {C_3}x^3 + ..... + {C_n}x^n$$(1 - x)^n = {C_0} - {C_1}x + {C_2}x^2 - {C_3

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$\frac{{3^n - (-1)^n}}{2}$

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(B) We know the binomial expansions:$(1 + x)^n = {C_0} + {C_1}x + {C_2}x^2 + {C_3}x^3 + ..... + {C_n}x^n$$(1 - x)^n = {C_0} - {C_1}x + {C_2}x^2 - {C_3}x^3 + ..... + (-1)^n{C_n}x^n$Subtracting the two equations:$(1 + x)^n - (1 - x)^n = 2({C_1}x + {C_3}x^3 + {C_5}x^5 + .....)$Dividing by $2$:$\frac{(1 + x)^n - (1 - x)^n}{2} = {C_1}x + {C_3}x^3 + {C_5}x^5 + .....$To get the required series $2.{C_1} + {2^3}.{C_3} + {2^5}.{C_5} + ....$,we substitute $x = 2$ in the expression ${C_1}x + {C_3}x^3 + {C_5}x^5 + .....$:$2.{C_1} + {2^3}.{C_3} + {2^5}.{C_5} + ..... = \frac{(1 + 2)^n - (1 - 2)^n}{2} = \frac{3^n - (-1)^n}{2}$

If ${C_0}, {C_1}, {C_2}, ......., {C_n}$ are the binomial coefficients,then $2.{C_1} + {2^3}.{C_3} + {2^5}.{C_5} + ....$ equals

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