The value of the sum left({ }^{n} C_{1}right) | vedclass

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(D) We know that the binomial expansion is given by: $(1+x)^{n} = { }^{n} C_{0}+{ }^{n} C_{1} x+{ }^{n} C_{2} x^{2}+\ldots+{ }^{n} C_{n} x^{n}$ and $(

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${ }^{2 n} C_{n}-1$

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(D) We know that the binomial expansion is given by: $(1+x)^{n} = { }^{n} C_{0}+{ }^{n} C_{1} x+{ }^{n} C_{2} x^{2}+\ldots+{ }^{n} C_{n} x^{n}$ and $(x+1)^{n} = { }^{n} C_{0} x^{n}+{ }^{n} C_{1} x^{n-1}+{ }^{n} C_{2} x^{n-2}+\ldots+{ }^{n} C_{n}$. Multiplying these two expressions, we get $(1+x)^{2n} = (1+x)^{n}(x+1)^{n}$. The coefficient of $x^{n}$ in the expansion of $(1+x)^{2n}$ is ${ }^{2n} C_{n}$. By multiplying the two series, the coefficient of $x^{n}$ is the sum of the products of corresponding coefficients: $({ }^{n} C_{0})^{2} + ({ }^{n} C_{1})^{2} + ({ }^{n} C_{2})^{2} + \ldots + ({ }^{n} C_{n})^{2} = { }^{2n} C_{n}$. The given sum starts from $({ }^{n} C_{1})^{2}$, so we subtract the first term $({ }^{n} C_{0})^{2} = 1$: $({ }^{n} C_{1})^{2} + ({ }^{n} C_{2})^{2} + \ldots + ({ }^{n} C_{n})^{2} = { }^{2n} C_{n} - ({ }^{n} C_{0})^{2} = { }^{2n} C_{n} - 1$.

The value of the sum $\left({ }^{n} C_{1}\right)^{2}+\left({ }^{n} C_{2}\right)^{2}+\left({ }^{n} C_{3}\right)^{2}+\ldots+\left({ }^{n} C_{n}\right)^{2}$ is

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