11th Class Mathematics Other Series Question Bank Critical Thinking

  • question_answer The sum of the series \[\frac{1}{1+{{1}^{2}}+{{1}^{4}}}+\frac{2}{1+{{2}^{2}}+{{2}^{4}}}+\frac{3}{1+{{3}^{2}}+{{3}^{4}}}+.........\] to \[n\] terms is

    A) \[\frac{n({{n}^{2}}+1)}{{{n}^{2}}+n+1}\]

    B) \[\frac{n(n+1)}{2({{n}^{2}}+n+1)}\]

    C) \[\frac{n({{n}^{2}}-1)}{2({{n}^{2}}+n+1)}\]

    D) None of these

    Correct Answer: B

    Solution :

    Let \[{{T}_{n}}\] be the \[{{n}^{th}}\] term of the series \[\frac{1}{1+{{1}^{2}}+{{1}^{4}}}+\frac{2}{1+{{2}^{2}}+{{2}^{4}}}+\frac{3}{1+{{3}^{2}}+{{3}^{4}}}+......\] Then \[{{T}_{n}}=\frac{n}{1+{{n}^{2}}+{{n}^{4}}}=\frac{n}{{{(1+{{n}^{2}})}^{2}}-{{n}^{2}}}\]            \[=\frac{n}{({{n}^{2}}+n+1)({{n}^{2}}-n+1)}\]           \[=\frac{1}{2}\left[ \frac{1}{{{n}^{2}}-n+1}-\frac{1}{{{n}^{2}}+n+1} \right]\]           \[=\frac{1}{2}\left[ \frac{1}{1+(n-1)n}-\frac{1}{1+n(n+1)} \right]\] Now  \[\sum\limits_{r=1}^{n}{{{T}_{r}}}=\frac{1}{2}\left[ \frac{1}{1}-\frac{1}{1+1.2} \right]+\frac{1}{2}\left[ \frac{1}{1+1.2}-\frac{1}{1+2.3} \right]\] \[+\frac{1}{2}\left[ \frac{1}{1+2.3}-\frac{1}{1+3.4} \right]+.......+\frac{1}{2}\left[ \frac{1}{1+(n-1)n}-\frac{1}{1+n(n+1)} \right]\] \[=\frac{1}{2}\left[ 1-\frac{1}{1+n(n+1)} \right]=\frac{n(n+1)}{2({{n}^{2}}+n+1)}\]. Trick: Checking for\[c\]. \[{{S}_{1}}=\frac{1}{3}\] and \[{{S}_{2}}=\frac{3}{7}\] which are given by (b).


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