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JEE Main & Advanced Sample Paper JEE Main Sample Paper-13

  • question_answer
    Direction: (Q. Nos. 89) For the following questions, the correct answers from the codes (a), (b), (c) and (d) defined as follows.
    We have, \[\sum{n=\frac{n(n+1)}{2},\,\,\sum{{{n}^{2}}}=\frac{n(n+1)\,(2n+1)}{6}}\] and \[\sum{{{n}^{3}}={{\left[ \frac{n(n+1)}{2} \right]}^{2}},\,\,n\in N}\] Statement I The sum of the series \[1+\,\,(1+2+4)\,+\,(4+6+9)\,+\,(9+12+16)\]  \[+...+\,\,(361+380\,+400)\] is 8000. Statement II \[\sum\limits_{k=1}^{n}{[{{k}^{3}}-{{(k-1)}^{3}}]={{n}^{3}}}\] for any natural number\[n\].

    A)  Statement I is true, Statement II is also true and Statement II is the correct explanation of Statement I.

    B)  Statement I is true, Statement II is also true and Statement II is not the correct explanation of Statement I.

    C)  Statement I is true, Statement II is false.

    D)  Statement I is false, Statement II is true.

    Correct Answer: A

    Solution :

     \[{{T}_{1}}=1,\]                \[{{T}_{2}}=7=(8-1)\] \[{{T}_{3}}=19=(27-8)\] ?????????? ?????????? \[{{T}_{n}}={{n}^{3}}-{{(n-1)}^{3}}\] \[\therefore \]  \[{{S}_{n}}=\Sigma \,\,{{T}_{n}}=({{n}^{3}}-{{n}^{3}}+1-3n+3{{n}^{2}})\] \[=\Sigma \,\left( 3{{n}^{2}}-3n+1 \right)\] \[=\frac{3n(n+1)\,(2n+1)}{6}\,-\frac{3n\,(n+1)}{2}+n\] \[=\frac{3n\,(n+1)}{6}\,(2n+1-3)+n\] \[=n(n+1)\,(n-1)+n\] \[=n[{{n}^{2}}-1+1]={{n}^{3}}\] \[\therefore \]  \[{{S}_{20}}={{(20)}^{3}}=8000\]


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