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JEE Main & Advanced Sample Paper JEE Main Sample Paper-8

  • question_answer
    The expansion of \[{{(1+x)}^{n}}\] has 3 consecutive terms with coefficients in the ratio 1 : 2 : 3 and can be written in the form\[^{n}{{C}_{k}}:{{\,}^{n}}{{C}_{k+1}}:{{\,}^{n}}{{C}_{k+2}}.\]The sum of all possible values of (n + k) is -

    A)  18                         

    B)  21

    C)  28                         

    D)  32

    Correct Answer: A

    Solution :

     \[\frac{^{n}{{C}_{k}}}{^{n}{{C}_{k+1}}}=\frac{1}{2}\Rightarrow \frac{n!}{k!(n-k)!}\frac{(k+1)!(n-k-1)!}{n!}=\frac{1}{2}\] or\[\frac{k+1}{n-k}=\frac{1}{2}\]\[2k+2=n-k\] \[n-3k=2\] ?(1) Similarly, \[\frac{^{n}{{C}_{k+1}}}{^{n}{{C}_{k+2}}}=\frac{2}{3}\] \[\frac{n!}{(k+1)!(n-k-1)!}.\frac{(k+2)!(n-k-2)!}{n!}=\frac{2}{3}\] \[\frac{k+2}{n-k-1}=\frac{2}{3}\]\[3+6=2n-2k-2\]\[2n-5=8\]?(2) From (1) and (2) n = 14 and k = 4 \[\therefore \]n + k = 18


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