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Question Number 93474 by Rio Michael last updated on 13/May/20

$$\mathrm{Q}.\mathrm{Prove}\mathrm{by}\mathrm{mathematical}\mathrm{induction}\mathrm{that}$$$$\underset{r=1}{\sum ^n}(4r+5)=2n^2+7n$$

Answered by eswar150933 last updated on 13/May/20

Answered by allizzwell23 last updated on 13/May/20

$${\mathit{P}}_1:4\left(1\right)+5=9=2{\left(1\right)}^2+7\left(1\right)\left(\mathbf{true}\right)$$$$\mathbf{Let}\mathit{n}=\mathit{k}$$$$\underset{r=1}{\sum ^n}(4r+5)=2k^2+7k$$$$\mathbf{Assume}\mathit{n}=\mathit{k}+1\mathbf{is}\mathbf{true}$$$$\underset{r=1}{\sum ^n}(4r+5)+(4(r+1)+5)$$$$=2{(k+1)}^2+7(k+1)$$$$\mathbf{To}\mathbf{prove}\mathbf{the}\mathbf{truth}\mathbf{of}\mathit{n}=\mathit{k}+1$$$$\underset{r=1}{\sum ^n}(4r+5)+(4(r+1)+5)$$$$=2k^2+7k+(4(k+1)+5)$$$$=2k^2+7k+4k+4+5$$$$=2k^2+4k+2+7k+7$$$$=2(k^2+2k+1)+7(k+1)$$$$=2{(k+1)}^2+7(k+1)$$$$\mathbf{DONE}!$$$$130520$$

Commented by Rio Michael last updated on 13/May/20

$$\mathrm{thanks}{\mathrm{y}}^{\prime }\mathrm{all}$$

Commented by allizzwell23 last updated on 13/May/20

$${\mathrm{you}}^{\prime }\mathrm{re}\mathrm{welcom}$$

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