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Question Number 121154 by zakirullah last updated on 05/Nov/20

Answered by Bird last updated on 05/Nov/20

$$letf\left(n\right)=u_n\Rightarrow u_n=6u_{m-1}-9u_{n-2}$$$$u_0=1and{u}_2=2$$$$e\Rightarrow u_{n+2}=6u_{n+1}-9u_n\Rightarrow$$$$u_{n+2}-6u_{n+1}+9u_n=0$$$$+\left(eq\right)\to r^2-6r+9=0\Rightarrow$$$${(r-3)}^2=0\Rightarrow r=3\left(double\right)\Rightarrow$$$$u_n=(an+b)3^n$$$$u_0=1=b$$$$u_2=(2a+b)3^2=2\Rightarrow$$$$2a+b=\frac{2}{9}\Rightarrow 2a=\frac{2}{9}-1=\frac{-7}{9}\Rightarrow$$$$a=-\frac{7}{18}\Rightarrow f\left(n\right)=(-\frac{7}{18}n+1)3^n$$$$$$

Commented by zakirullah last updated on 05/Nov/20

$$thsnkssir!pleasesolveall?$$

Answered by TANMAY PANACEA last updated on 05/Nov/20

$$2)f\left(n\right)=f(n-1)+(n-1)$$$$f\left(2\right)=1=f\left(0\right)+1$$$$f\left(0\right)=0f\left(2\right)=1$$$$f\left(n\right)=\frac{n(n-1)}{2}$$$$$$$$$$

Commented by zakirullah last updated on 05/Nov/20

$$thankssirG$$

Commented by TANMAY PANACEA last updated on 06/Nov/20

$$mostwelcome$$

Answered by Bird last updated on 05/Nov/20

$$f\left(n\right)=f(n-1)+n-1\Rightarrow$$$$u_n=u_{n-1}+n-1\Rightarrow u_{n+1}=u_n+n\Rightarrow$$$$u_{n+1}-u_n=n\Rightarrow \sum _{k=1}^{n-1}(u_{k+1}-u_k)$$$$=\sum _{k=1}^{n-1}k=\frac{(n-1)n}{2}\Rightarrow$$$$u_2-u_1+u_3-u_2+....+u_n-u_{n-1}$$$$=\frac{n(n-1)}{2}\Rightarrow u_n-u_1=\frac{n(n-1)}{2}$$$$u_n=\frac{n(n-1)}{2}+u_1$$$$u_2=1+u_1=1\Rightarrow u_1=0\Rightarrow u_n=\frac{n(n-1)}{2}$$

Answered by Bird last updated on 05/Nov/20

$$f\left(n\right)=3f(n-1)\Rightarrow u_n=3u_{n-1}\Rightarrow$$$$\frac{u_n}{u_{n-1}}=3\Rightarrow \prod _{k=1}^n\frac{u_k}{u_{k-1}}=3^n\Rightarrow$$$$\frac{u_1}{u_0}.\frac{u_2}{u_1}......\frac{u_n}{u_{n-1}}=3^n\Rightarrow u_n=3^n{u}_0$$$$\Rightarrow u_n=2\times 3^n$$

Commented by zakirullah last updated on 05/Nov/20

$$thankssomuch$$

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