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JEE Main & Advanced Sample Paper JEE Main Sample Paper-22

  • question_answer
    The 7th  term of the series\[1+\frac{1}{(1+3)}{{(1+2)}^{2}}\frac{1}{(1+3+5)}{{(1+2+3)}^{2}}+........,\]equal to

    A)  16        

    B) 18

    C) 20                                

    D) 22

    Correct Answer: A

    Solution :

                            \[{{T}_{r}}=\frac{{{(1+2+........+r)}^{2}}}{{{r}^{2}}}\,=\frac{1}{{{r}^{2}}}\,{{\left( \frac{r(r+1)}{2} \right)}^{2}}\] \[=\frac{{{r}^{2}}+2r+1}{4}\] \[\therefore \,\,\,\,\,\,\,{{T}_{7}}=16\]


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