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Question Number 83242 by peter frank last updated on 29/Feb/20

Commented by john santu last updated on 29/Feb/20

$$\left(4\mathrm{b}\right)\underset{\mathrm{r}=1}{\sum ^{\mathrm{n}}}\mathrm{r}(\mathrm{r}+1)=\underset{\mathrm{r}=1}{\sum ^{\mathrm{n}}}{\mathrm{r}}^2+\underset{\mathrm{r}=1}{\sum ^{\mathrm{n}}}\mathrm{r}$$$$=\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6}+\frac{\mathrm{n}(\mathrm{n}+1)}{2}$$$$=\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)+3\mathrm{n}(\mathrm{n}+1)}{6}$$$$=\frac{\mathrm{n}(\mathrm{n}+1)[2\mathrm{n}+1+3]}{6}$$$$=\frac{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)}{3}$$

Commented by peter frank last updated on 01/Mar/20

$$4a,c$$

Commented by peter frank last updated on 01/Mar/20

$$thankyou$$

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