If (1+x) n=C0+C1x+C2x2+...+Cnxn, then 0≤i≤ ∑ j≤n∑ (Ci+Cj) 2=

∑0≤i≤ ∑j≤n (Ci+Cj) 2 i=0,1,2,...,(n−1);j=1,2,3,...,n and i<j=n(C02+C12+...+Cn2)+2∑∑CiCj 0≤i≤j≤n=n.2nCn+[(C0+C1+...+Cn) 2−(C02+C12+...+Cn) ]=n.2nCn+(2n) 2−2nCn=(n−1).2nCn+22n

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If { ( 1+x ) }^{ n }={ C }_{ 0 }+{ C }_{ 1 }x+{ C }_{ 2 }{ x

Question

If $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + ... + C_{n} x^{n}$ , then $0 \leq i \leq \sum j \leq n \sum (C_{i} + C_{j})^{2} =$

(n - 1) ._{2 n} C_{n} + 2^{2 n}

n ._{2 n} C_{n} + 2^{2 n}

(n + 1) ._{2 n} C_{n} + 2^{2 n}

None of these

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(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + ... + C_{n} x^{n}

, then

0 \leq i \leq \sum j \leq n \sum (C_{i} + C_{j})^{2} =

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Text solutionVerified

\sum_{0 \leq i \leq} \sum_{j \leq n} (C_{i} + C_{j})^{2} i = 0, 1, 2, ..., (n - 1); j = 1, 2, 3, ..., n

and

i < j

= n (C_{0}^{2} + C_{1}^{2} + ... + C_{n}^{2}) + 2 \sum \sum C_{i} C_{j} 0 \leq i \leq j \leq n = n .^{2 n} C_{n} + [(C_{0} + C_{1} + ... + C_{n})^{2} - (C_{0}^{2} + C_{1}^{2} + ... + C_{n})] = n .^{2 n} C_{n} + (2^{n})^{2} -^{2 n} C_{n} = (n - 1) .^{2 n} C_{n} + 2^{2 n}

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Question Text	If $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + ... + C_{n} x^{n}$ , then $0 \leq i \leq \sum j \leq n \sum (C_{i} + C_{j})^{2} =$
Answer Type	Text solution:1
Upvotes	150

If (1+x) n=C0​+C1​x+C2​x2+...+Cn​xn, then 0≤i≤ ∑ ​j≤n∑ ​(Ci​+Cj​) 2=

Text solutionVerified

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If $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + ... + C_{n} x^{n}$ , then $0 \leq i \leq \sum j \leq n \sum (C_{i} + C_{j})^{2} =$