The correct option is A True ( 1+x )^n=C_0+C_1x+C_2x^2+...+C_nx^n #160; #160; #160; #160; #160; #160; #160; #160; #160; (i) ( x+1 ...

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Implicit Differentiation

C02+C12+C22+....

Question

${C_{0}}^{2} + {C_{1}}^{2} + {C_{2}}^{2} + . . . + {C_{n}}^{2} =^{2 n} C_{n}$

True

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False

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Solution

The correct option is A True
${(1 + x)}^{n} = C_{0} + C_{1} x + C_{2} x^{2} + . . . + C_{n} x^{n}$ (i)
${(x + 1)}^{n} = C_{0} x^{n} + C_{1} x^{n - 1} C_{2} x^{n - 2} + . . . + C_{n} x^{n}$ (ii)
Multiplying (i) and (ii)
$(C_{0} + C_{1} x + C_{2} x^{2} + . . . + C_{n} x^{n}) (C_{0} x^{n} + C_{1} x^{n - 1} C_{2} x^{n - 2} + . . . + C_{n} x^{n}) = {(1 + x)}^{2 n}$
Comparing coefficient of $x^{n}$
${C_{0}}^{2} + {C_{1}}^{2} + {C_{2}}^{2} + . . . + {C_{n}}^{2} =^{2 n} C_{n}$

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Similar questions

Find the value of $C_{0}^{2} + C_{1}^{2} + C_{2}^{2} . . . . . C_{n}^{2}$ , where $C_{r} =^{n} C_{r}$

1. {C_{0}}^{2} + 3. {C_{1}}^{2} + 5. {C_{2}}^{2} + . . . + (2 n + 1) . {C_{n}}^{2} = 2 n .^{2 n - 1} C_{n} +^{2 n} C_{n} .

^{2 n} C_{n} = C_{0}^{2} + C_{1}^{2} + C_{2}^{2} + C_{3}^{2} + \dots \dots + C_{n}^{2}

^{2 n} C_{n} =

x

(1 + x)^{n} {(1 + \frac{1}{x})}^{n}

^{2 n} C_{n} = \frac{1.3.5.7 \dots \dots (2 n - 1)}{n!}

C_{0}^{2} + 3. C_{1}^{2} + 5. C_{2}^{2} + \dots \dots + (2 n + 1) . C_{n}^{2} =

C_{0}^{2} - C_{1}^{2} + C_{2}^{2} - . . . . . . + {(- 1)}^{n} \times C_{n}^{2}

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