Question

दर्शाइए कि: ${\displaystyle \frac{1\times 2^2+2\times 3^2+....+ \text{n}\times {(\text{n}+1)}^2}{1^2\times 2+2^2\times 3+....+ {\text{n}}^2\times (\text{n}+1)}=\frac{3\text{n}+5}{3\text{n}+1}}$

Answer 1

अंश का nवाँ पद = n(n + 1)² = n(n² + 2n + 1)

= n³ + 2n² + n

अंश के n पदों का योग S₁ = ${\displaystyle \sum {\text{n}}^3+2\sum {\text{n}}^2+\sum \text{n}}$

= ${\displaystyle \frac{{\text{n}}^2{(\text{n}+1)}^2}{4}+\frac{2\text{n}(\text{n}+1)(2\text{n}+1)}{6}+\frac{\text{n}(\text{n}+1)}{2}}$

= ${\displaystyle \frac{\text{n}(\text{n}+1)}{12}[3\text{n}(\text{n}+1)+4(2\text{n}+1)+6]}$

= ${\displaystyle \frac{\text{n}(\text{n}+1)}{12}[3{\text{n}}^2+3\text{n}+8\text{n}+4+6]}$

= ${\displaystyle \frac{\text{n}(\text{n}+1)}{12}[3{\text{n}}^2+11\text{n}+10]}$

= ${\displaystyle \frac{\text{n}(\text{n}+1)(\text{n}+2)(3\text{n}+5)}{12}}$

हर का nवाँ पद = n² (n + 1) = n³ + n²

हर के n पदों का योग S₂ = ${\displaystyle \sum {\text{n}}^3+\sum {\text{n}}^2}$

= ${\displaystyle \frac{{\text{n}}^2{(\text{n}+1)}^2}{4}+\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}}$ 

= ${\displaystyle \frac{\text{n}(\text{n}+1)}{12}[3\text{n}(\text{n}+1)+2(2\text{n}+1)]}$

= ${\displaystyle \frac{\text{n}(\text{n}+1)}{12}[3{\text{n}}^2+3\text{n}+4\text{n}+2]}$

= ${\displaystyle \frac{\text{n}(\text{n}+1)}{12}[3{\text{n}}^2+7\text{n}+2]}$

= ${\displaystyle \frac{\text{n}(\text{n}+1)(\text{n}+2)(3\text{n}+1)}{12}}$

${\displaystyle \frac{{\text{S}}_1}{{\text{S}}_2}=\frac{\frac{\text{n}(\text{n}+1)(\text{n}+2)(3\text{n}+5)}{12}}{\frac{\text{n}(\text{n}+1)(\text{n}+2)(3\text{n}+1)}{12}}}$

= ${\displaystyle \frac{3\text{n}+5}{3\text{n}+1}}$