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Given:: Sn=∑r=1ntr=n(2n2+9n+13)6Given :: \ \displaystyle S_{n}=\sum_{r=1}^n t_{r}= \frac{n(2n^{2}+9n+13)}{6}Given:: Sn=r=1∑ntr=6n(2n2+9n+13) Find∑r=1∞1r.trFind \quad \displaystyle \sum_{r=1}^\infty \frac{1}{r.\sqrt{t_{r}}}Findr=1∑∞r.tr1
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