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Question Number 146672 by mathdanisur last updated on 14/Jul/21

Answered by Olaf_Thorendsen last updated on 15/Jul/21

$$\mathrm{S}=\frac{\underset{n=1}{\sum ^x}n{(n+1)}^2}{\underset{n=1}{\sum ^x}{n}^2(n+1)}$$$$\mathrm{S}=\frac{\underset{n=1}{\sum ^x}{n}^3+2n^2+n}{\underset{n=1}{\sum ^x}{n}^3+n^2}$$$$\mathrm{S}=\frac{\frac{x^2{(x+1)}^2}{4}+2.\frac{x(x+1)(2x+1)}{6}+\frac{x(x+1)}{2}}{\frac{x^2{(x+1)}^2}{4}+\frac{x(x+1)(2x+1)}{6}}$$$$\mathrm{S}=\frac{\frac{x(x+1)}{4}+\frac{(2x+1)}{3}+\frac{1}{2}}{\frac{x(x+1)}{4}+\frac{(2x+1)}{6}}$$$$\mathrm{S}=\frac{3x(x+1)+4(2x+1)+6}{3x(x+1)+2(2x+1)}$$$$\mathrm{S}=\frac{3x^2+11x+10}{3x^2+7x+2}=\frac{5}{4}$$$$12x^2+44x+40=15x^2+35x+10$$$$3x^2-9x-30=0$$$$3(x+2)(x-5)=0\Rightarrow x=5$$

Commented by mathdanisur last updated on 15/Jul/21

$$Ser,ture.?$$

Commented by mathdanisur last updated on 14/Jul/21

$$ThankyouSer,answer:x=5$$

Commented by bramlexs22 last updated on 15/Jul/21

$$\mathrm{your}\mathrm{answer}\mathrm{x}=5\mathrm{is}\mathrm{wrong}$$

Commented by som(math1967) last updated on 15/Jul/21

$$sir,3rdline$$$$\underset{n=1}{\sum ^x}{n}^3+n^{2}shouldbe\frac{x^2{(x+1)}^2}{4}+\frac{x(x+1)(2x+1)}{6}$$

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