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  • <img alt="\text { If }(1+x)^n=C_0+C_1 x+C_2 x^2+\ldots \ldots \ldots+C_n x^2 \text {, then }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Ctext%20%7B%20If%20%7D%281&plus;x%29%5En%3DC

\text { If }(1+x)^n=C_0+C_1 x+C_2 x^2+\ldots \ldots \ldots+C_n x^2 \text {, then } C_0^2+C_1^2+C_2^2+C_3^2+\ldots \ldots+C_n^2=

Option: 1

\frac{n !}{n ! n !}


Option: 2

\frac{(2 n) !}{n ! n !}


Option: 3

\frac{(2 n) !}{n !}


Option: 4

None of these


Answers (1)

best_answer

(1+x)^n=C_0+C_1 x+C_2 x^2+\ldots . .+C_n x^n....(i)\text { and }\left(1+\frac{1}{x}\right)^n=C_0+C_1 \frac{1}{x}+C_2\left(\frac{1}{x}\right)^2+\ldots . .+C_n\left(\frac{1}{x}\right)^n....(ii)

If we multiply (i) and (ii), we get

C_0^2+C_1^2+C_2^2+\ldots . .+C_n^2

is the term independent of x and hence it is equal to the term independent of x  in the product (1+x)^n\left(1+\frac{1}{x}\right)^n \text { or in } \frac{1}{x^n}(1+x)^{2 n}  or term containing x^n \text { in }(1+x)^{2 n} . Clearly the coefficient of xn in (1+x)^{2 n} \text { is } T_{n+1} and equal to { }^{2 n} C_n=\frac{(2 n) !}{n ! n !}

Posted by

HARSH KANKARIA

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