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Question Number 73346 by TawaTawa last updated on 10/Nov/19

Commented by mathmax by abdo last updated on 10/Nov/19

$$a)\sum _{k=0}^n{(2+3k)}^2=4+\sum _{k=1}^n(4+12k+9k^2)$$$$=4+4n+12\sum _{k=1}^{n}k+9\sum _{k=1}^n{k}^2$$$$=4n+4+12\frac{n(n+1)}{2}+9\frac{n(n+1)(2n+1)}{6}$$$$=4n+4+6n(n+1)+\frac{3}{2}n(n+1)(2n+1).$$

Commented by mathmax by abdo last updated on 10/Nov/19

$$b)\sum _{k=0}^n{C}_n^k{3}^k=\sum _{k=0}^n{C}_n^k{3}^k\times 1^{n-k}={(3+1)}^n=4^n$$

Commented by mathmax by abdo last updated on 10/Nov/19

$$\sum _{i=0}^{n-1}{C}_n^{i+1}{(-1)}^i{=}_{i+1=k}\sum _{k=1}^n{C}_n^k{(-1)}^{k-1}=-\sum _{k=1}^n{C}_n^k{(-1)}^k$$$$=-(\sum _{k=0}^n{C}_n^k{(-1)}^k\times 1^{n-k}-1)=-(0-1)=1$$

Commented by TawaTawa last updated on 10/Nov/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by mathmax by abdo last updated on 10/Nov/19

$$youarewelcome.$$

Answered by JDamian last updated on 10/Nov/19

$$\left(b\right)\underset{k=0}{\sum ^n}{3}^k\left(\begin{array}{c}n\\ k\end{array}\right)=\underset{k=0}{\sum ^n}{3}^k{1}^{n-k}\left(\begin{array}{c}n\\ k\end{array}\right)={(3+1)}^n=4^n$$

Commented by TawaTawa last updated on 10/Nov/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by JDamian last updated on 10/Nov/19

$$\left(a\right)\underset{k=0}{\sum ^n}{(2+3k)}^2=$$$$=\underset{k=0}{\sum ^n}(4+12k+9k^2)=$$$$=\underset{k=0}{\sum ^n}4+12\underset{k=0}{\sum ^n}k+9\underset{k=0}{\sum ^n}{k}^2=$$$$=4(n+1)+12\frac{(n+1)n}{2}+9\frac{n(n+1)(2n+1)}{6}=$$

Commented by TawaTawa last updated on 10/Nov/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by mr W last updated on 10/Nov/19

$$\underset{i=0}{\sum ^{n-1}}{(-1)}^i\left({}_{i+1}^n\right)$$$$=\underset{i=1}{\sum ^n}{(-1)}^{k-1}\left({}_k^n\right)$$$$=-\underset{i=1}{\sum ^n}{(-1)}^k\left({}_k^n\right)$$$$=-\underset{i=0}{\sum ^n}{(-1)}^k\left({}_k^n\right)+1$$$$=-{(1-1)}^n+1$$$$=1$$

Commented by TawaTawa last updated on 10/Nov/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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