જો $n$ એ $1$ કરતા મોટો ધન પૂર્ણાંક હોય,તો $3({ }^n C_0) - 8({ }^n C_1) + 13({ }^n C_2) - 18({ }^n C_3) + \ldots$ $(n+1)$ પદો સુધી $=$

  • A
    -$5$
  • B
    $\frac{2^{n+1}-1}{n}$
  • C
    $\frac{2^n-1}{2}$
  • D
    $0$

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જો $3 \times { }^5 C_0 + 8 \times { }^5 C_1 + 13 \times { }^5 C_2 + 18 \times { }^5 C_3 + 23 \times { }^5 C_4 + 28 \times { }^5 C_5 = k \times 2^4$ હોય,તો $k=$

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