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JEE Main & Advanced JEE Main Paper Phase-I Held on 07-1-2020 Morning

  • question_answer
    The greatest positive integer k, for which \[{{49}^{k}}+1\] is a factor of the sum \[{{49}^{125}}+{{49}^{124}}+.....+{{49}^{2}}+49+1,\] is [JEE MAIN Held on 07-01-2020 Morning]

    A) 65         

    B) 60

    C) 32                    

    D) 63

    Correct Answer: D

    Solution :

    [d] \[1+49+{{49}^{2}}+.....+{{(49)}^{125}}\] \[\frac{{{(49)}^{126}}-1}{48}=\frac{\left( {{(49)}^{63}}-1 \right)\left( {{(49)}^{63}}+1 \right)}{48}\] Greatest value of k is 63.


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