If the sum 3 1^2+ 5 1^2+2^2+ 7 1^2+2^2+3^2+.....+ up to 20 terms is equal to k 21, then k is equal to

$${S}_{20}=\frac{3}{{1}^{2}}+\frac{5}{{1}^{2}+{2}^{2}}+\frac{7}{{1}^{2}+{2}^{2}+{3}^{2}}+....+\mathrm{upto}20terms$$S20​=123​+12+225​+12+22+327​+....+upto20terms

$$=\Sigma _{r=1}^{20}\frac{(2r+1)}{{1}^{2}+{2}^{2}+{3}^{2}+....{r}^{2}}$$=Σr=120​12+22+32+....r2(2r+1)​

$$=\Sigma _{r=1}^{20}\frac{2r+1}{\frac{r}{6}(r+1)(2r+1)}$$=Σr=120​6r​(r+1)(2r+1)2r+1​

$$=6\Sigma _{r=1}^{20}\frac{1}{r(r+1)}$$=6Σr=120​r(r+1)1​

$$=6\Sigma _{r=1}^{20}(\frac{1}{r}-\frac{1}{r+1})$$=6Σr=120​(r1​−r+11​)

$$=6{[\frac{1}{1}-\frac{1}{2}]+[\frac{1}{2}-\frac{1}{3}]+[\frac{1}{3}-\frac{1}{4}]+....+[\frac{1}{20}-\frac{1}{21}]}$$=6[11​−21​]+[21​−31​]+[31​−41​]+....+[201​−211​]

$$=6(1-\frac{1}{21})$$=6(1−211​)

$$=\frac{6\times 20}{21}$$=216×20​

$$=\frac{120}{21}$$=21120​

$$=\frac{k}{21}$$=21k​

$$k=120$$k=120