Dear student Let S=11+12+14+21+22+24+31+32+34+ +991+992+994So, Tr=r1+r2+r4=rr4+2r2+1-r2=rr2+12-r2=2r2r2+1-rr2+1+r=12r2+r+1-r2+1-rr2+1-rr2+1+r=121r2+1-r-1r2+1+r=∑r=199Tr=12∑r=1991r2+1-r-1r2+1+r=121-19801+100=121-19901=49509901=0 4999So correct option is (4) Regards

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Dear student

Let S = \frac{1}{1 + 1^{2} + 1^{4}} + \frac{2}{1 + 2^{2} + 2^{4}} + \frac{3}{1 + 3^{2} + 3^{4}} + . . . + \frac{99}{1 + 99^{2} + 99^{4}} So, T_{r} = \frac{r}{1 + r^{2} + r^{4}} = \frac{r}{r^{4} + 2 r^{2} + 1 - r^{2}} = \frac{r}{{(r^{2} + 1)}^{2} - r^{2}} = \frac{2 r}{2 (r^{2} + 1 - r) (r^{2} + 1 + r)} = \frac{1}{2} [\frac{(r^{2} + r + 1) - (r^{2} + 1 - r)}{(r^{2} + 1 - r) (r^{2} + 1 + r)}] = \frac{1}{2} [\frac{1}{(r^{2} + 1 - r)} - \frac{1}{(r^{2} + 1 + r)}] = \sum_{r = 1}^{99} T_{r} = \frac{1}{2} \sum_{r = 1}^{99} (\frac{1}{r^{2} + 1 - r} - \frac{1}{r^{2} + 1 + r}) = \frac{1}{2} [1 - \frac{1}{9801 + 100}] = \frac{1}{2} [1 - \frac{1}{9901}] = \frac{4950}{9901} = 0.4999 So correct option is (4)

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