If {left( {1 + x} right)^n} = {c_0} + {c | vedclass

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Question

$(1+x)^{n}=C_{0}+C_{1} x+C_{2} x^{2}+\ldots . C_{n} x^{n}$

$\mathrm{x} \rightarrow \mathrm{x}^{2}$

$\left(1+x^{2}\right)^{n}=C_{0}+C_{1} x^{2}+C

Vedclass · Accepted Answer

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Vedclass · Answer

$(1+x)^{n}=C_{0}+C_{1} x+C_{2} x^{2}+\ldots . C_{n} x^{n}$

$\mathrm{x} \rightarrow \mathrm{x}^{2}$

$\left(1+x^{2}\right)^{n}=C_{0}+C_{1} x^{2}+C_{2} x^{4}+\ldots . C_{n} x^{2 n}$

Multiplying $x,$

$\mathrm{x}\left(1+\mathrm{x}^{2}\right)^{\mathrm{n}}=\mathrm{C}_{0} \mathrm{x}+\mathrm{C}_{1} \mathrm{x}^{3}+\mathrm{C}_{2} \mathrm{x}^{5}+\ldots . \mathrm{C}_{\mathrm{n}} \mathrm{x}^{2 \mathrm{n}+1}$

Differentiating both sides w.r.t. $\mathrm{x}$, then

$x.n.{\left( {1 + {x^2}} \right)^{n - 1}} \cdot 2x + {\left( {1 + {x^2}} \right)^n} = {C_0} + 3{C_1}{x^2} + 5{C_2}{x^4}$

$ +  \ldots  \ldots  + (2n + 1){C_n}{x^{2n}}$

Put $x=i$ in both sides

$0+0=\mathrm{C}_{0}-3 \mathrm{C}_{1}+5 \mathrm{C}_{2}-\ldots \ldots .(-1)^{\mathrm{n}}(2 \mathrm{n}+1) \mathrm{C}_{\mathrm{n}}=0$

If ${\left( {1 + x} \right)^n} = {c_0} + {c_1}x + {c_2}{x^2} + {c_3}{x^3} + ...... + {c_n}{x^n}$ , then the value of ${c_0} - 3{c_1} + 5{c_2} - ........ + {( - 1)^n}\,(2n + 1){c_n}$ is

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