11+12+14+21+22+24+31+32+34+...... (n terms ) =

Tr=12{11−r+r2−11+r+r2} ∑r=1nTr=12∑r=1n{11−r+r2−11+r+r2} 12{(1−13)+(13−17)+(17−113)+...+(11−n+n2−11+n+n2)} =12(1−11+n+n2)=n(n+1)2(n2+n+1)

11+12+14+21+22+24+31+32+34+...... (n terms ) =

Tr=12{11−r+r2−11+r+r2} ∑r=1nTr=12∑r=1n{11−r+r2−11+r+r2} 12{(1−13)+(13−17)+(17−113)+...+(11−n+n2−11+n+n2)} =12(1−11+n+n2)=n(n+1)2(n2+n+1)

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$\frac{1}{1 + 1^{2} + 1^{4}} + \frac{2}{1 + 2^{2} + 2^{4}} + \frac{3}{1 + 3^{2} + 3^{4}} + ......$ (n terms ) =

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$\frac{n (n + 1)}{2 (n^{2} + n + 1)}$

$\frac{n (n - 1)}{2 (n^{2} - n + 1)}$

$\frac{n (n + 1)}{3 (n^{2} + n - 1)}$

$\frac{n (n - 1)}{n^{2} + n + 1}$

answer is A.

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Detailed Solution

$T_{r} = \frac{1}{2} {\frac{1}{1 - r + r^{2}} - \frac{1}{1 + r + r^{2}}} \sum_{r = 1}^{n} T_{r} = \frac{1}{2} \sum_{r = 1}^{n} {\frac{1}{1 - r + r^{2}} - \frac{1}{1 + r + r^{2}}} \frac{1}{2} {(1 - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{7}) + (\frac{1}{7} - \frac{1}{13}) + ... + (\frac{1}{1 - n + n^{2}} - \frac{1}{1 + n + n^{2}})} = \frac{1}{2} (1 - \frac{1}{1 + n + n^{2}}) = \frac{n (n + 1)}{2 (n^{2} + n + 1)}$