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Question Number 77953 by pete last updated on 12/Jan/20

$$\mathrm{Given}\mathrm{that}\underset{\mathrm{r}=0}{\sum ^4}6\mathrm{r}=2\underset{\mathrm{r}=1}{\sum ^{\mathrm{n}}}5\mathrm{r},\mathrm{work}\mathrm{out}\mathrm{the}\mathrm{value}$$$$\mathrm{of}\mathrm{n}.$$

Answered by john santu last updated on 12/Jan/20

$$\underset{0}{\sum ^4}\left(\frac{3}{5}\right)r=\underset{r=1}{\sum ^n}r$$$$\frac{3}{5}(0+1+2+3+4)=\frac{n}{2}(n+1)$$$$12=n^2+n\iff n^2-n-12=0$$$$n=4,n=-3\left(nosolution\right)$$

Commented by pete last updated on 12/Jan/20

$$\mathrm{thanks}\mathrm{but}\mathrm{i}\mathrm{dont}\mathrm{understand}\mathrm{please}.$$$$\mathrm{kindly}\mathrm{do}\mathrm{some}\mathrm{explaining}\mathrm{for}\mathrm{me}.$$

Answered by JDamian last updated on 12/Jan/20

$$3\underset{\mathrm{r}=0}{\sum ^4}\mathrm{r}=5\underset{\mathrm{r}=1}{\sum ^{\mathrm{n}}}\mathrm{r}$$$$3\frac{0+4}{2}5=5\frac{n+1}{2}n$$$$12=n(n+1)$$$$n^2+n-12=0$$

Commented by pete last updated on 12/Jan/20

$$\mathrm{thanks}\mathrm{sir}$$

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