If 1-xn-C0-C1 x-C2 x2-Cn xn- then -0 - r -s - n-nr-sCr-Cs-Cr Cs is equal toA- n2- 2n-n-2-22 n-2 n Cn-B- n2- 2n-2-22 n-2 n Cn-C- n2- 2n-2-22 n-2 n Cn-D- None of these

The correct option is B n^2 · 2^n+n2[2^2n ^2nC_n]We have, 0 ≤ r lt;s ≤ n∑∑ (r+s)(C_r+ C_s+C_rC_s) =0 ≤ r lt;s ≤ n∑∑ (r+s)(C_r +C_s)+0 ≤ r lt;s ≤ n∑∑ (r ...

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Byju's Answer

Standard XII

Mathematics

Sum of Coefficients of All Terms

If 1+xn=C0+C1...

Question

If $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n},$ then $\sum \sum 0 \leq r < s \leq n (r + s) (C_{r} + C_{s} + C_{r} C_{s})$ is equal to

n^{2} \cdot 2^{n} - \frac{n}{2} [2^{2 n} -^{2 n} C_{n}]

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n^{2} \cdot 2^{n} + \frac{n}{2} [2^{2 n} -^{2 n} C_{n}]

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n^{2} \cdot 2^{n} + \frac{n}{2} [2^{2 n} +^{2 n} C_{n}]

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None of these

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Solution

The correct option is B $n^{2} \cdot 2^{n} + \frac{n}{2} [2^{2 n} -^{2 n} C_{n}]$
We have,
$\sum \sum 0 \leq r < s \leq n (r + s) (C_{r} + C_{s} + C_{r} C_{s}) = \sum \sum 0 \leq r < s \leq n (r + s) (C_{r} + C_{s}) + \sum \sum 0 \leq r < s \leq n (r + s) C_{r} C_{s}$

$= n^{2} \cdot 2^{n} + \frac{n}{2} [2^{2 n} -^{2 n} C_{n}]$

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Similar questions

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(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n},

then the value of

\sum \sum 0 \leq r < s \leq n (r \cdot s) C_{r} C_{s}

Q. If

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\frac{C_{0}}{1} + \frac{C_{2}}{3} + \frac{C_{4}}{5} + . . . . =

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Q. If

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

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If $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + . . . . . . . + C_{n} x^{n}$ , then $\frac{C_{1}}{C_{0}} + \frac{2 C_{2}}{C_{1}} + \frac{3 C_{3}}{C_{2}} + . . . . . . . . + \frac{n C_{n}}{C_{n - 1}} =$

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If (1+x)n=C0+C1x+C2x2+…+Cnxn, then ∑∑0≤r<s≤n(r+s)(Cr+Cs+CrCs) is equal to

The correct option is B n2⋅2n+n2[22n− 2nCn]We have, ∑∑0≤r<s≤n(r+s)(Cr+Cs+CrCs)=∑∑0≤r<s≤n(r+s)(Cr+Cs)+∑∑0≤r<s≤n(r+s)CrCs =n2⋅2n+n2[22n− 2nCn]

If $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n},$ then $\sum \sum 0 \leq r < s \leq n (r + s) (C_{r} + C_{s} + C_{r} C_{s})$ is equal to