Let r th term of a series be given by T r - r -1-3 r 2- r 4- Then -2 - r -1- T r is

T_r= r1 3r^2+r^4=r(r^2 1)^2 r^2 =12[1r^2 1 r 1r^2 1+r] So, ∑^n_r=1T_r =12[ 1 1] +12[1 15] +12[15 111] +12[111 119] + ⋮ ⋮ + 12[1n^2 1 n 1n^2 ...

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Vn Method

Let r th term...

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Let $r^{th}$ term of a series be given by $T_{r} = \frac{r}{1 - 3 r^{2} + r^{4}} .$ Then $- 2 \infty \sum r = 1 T_{r}$ is

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r^{th}

T_{r} = \frac{r}{1 - 3 r^{2} + r^{4}} .

- 2 \infty \sum r = 1 T_{r}

r^{t h}

T_{r} = \frac{r}{1 - 3 r^{2} + r^{4}} .

lim n \to \infty \sum_{r = 1}^{n} T_{r}

t_{r}

t_{r} = \frac{r}{1 + r^{2} + r^{4}}

{lim}_{n \to \infty} \sum_{r = 1}^{n} t_{r}

r^{t h}

T_{r}

T_{r} = \frac{r}{1 + r^{2} + r^{4}}

{lim}_{n \to \infty} n \sum r = 1 T_{r}

r t h

t_{r}

t_{r} = \frac{r}{1 + r^{2} + r^{4}}

lim n \to \infty n \sum r = 1 t_{r}

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