Question-196324

Question Number 196324 by sonukgindia last updated on 22/Aug/23

Answered by sniper237 last updated on 22/Aug/23

a(22)−a(21)=3×21−1 a(21)−a(20)=3×20−1 ... a(3)−a(2)=3×2−1 a(2)−a(1)=3×1−1 when adding member to member a(22)−a(1)=3×((21×22)/2)−21

$${a}\left(\mathrm{22}\right)−{a}\left(\mathrm{21}\right)=\mathrm{3}×\mathrm{21}−\mathrm{1} \\ $$$${a}\left(\mathrm{21}\right)−{a}\left(\mathrm{20}\right)=\mathrm{3}×\mathrm{20}−\mathrm{1} \\ $$$$… \\ $$$${a}\left(\mathrm{3}\right)−{a}\left(\mathrm{2}\right)=\mathrm{3}×\mathrm{2}−\mathrm{1} \\ $$$${a}\left(\mathrm{2}\right)−{a}\left(\mathrm{1}\right)=\mathrm{3}×\mathrm{1}−\mathrm{1} \\ $$$${when}\:{adding}\:{member}\:{to}\:{member} \\ $$$${a}\left(\mathrm{22}\right)−{a}\left(\mathrm{1}\right)=\mathrm{3}×\frac{\mathrm{21}×\mathrm{22}}{\mathrm{2}}−\mathrm{21} \\ $$

Answered by mr W last updated on 22/Aug/23

a(k+1)=a(k)+3k−1 Σ_(k=1) ^n a(k+1)=Σ_(k=1) ^n a(k)+Σ_(k=1) ^n (3k−1) Σ_(k=2) ^(n+1) a(k)=Σ_(k=1) ^n a(k)+Σ_(k=1) ^n (3k−1) a(n+1)−a(1)+Σ_(k=1) ^n a(k)=Σ_(k=1) ^n a(k)+Σ_(k=1) ^n (3k−1) a(n+1)−a(1)=Σ_(k=1) ^n (3k−1) a(n+1)−8=3×((n(n+1))/2)−n a(n+1)=((n(3n+1))/2)+8 ⇒a(n)=(((n−1)(3n−2))/2)+8 ⇒a(22)=((21×64)/2)+8=680

$${a}\left({k}+\mathrm{1}\right)={a}\left({k}\right)+\mathrm{3}{k}−\mathrm{1} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{a}\left({k}+\mathrm{1}\right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{a}\left({k}\right)+\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{3}{k}−\mathrm{1}\right) \\ $$$$\underset{{k}=\mathrm{2}} {\overset{{n}+\mathrm{1}} {\sum}}{a}\left({k}\right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{a}\left({k}\right)+\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{3}{k}−\mathrm{1}\right) \\ $$$${a}\left({n}+\mathrm{1}\right)−{a}\left(\mathrm{1}\right)+\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{a}\left({k}\right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{a}\left({k}\right)+\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{3}{k}−\mathrm{1}\right) \\ $$$${a}\left({n}+\mathrm{1}\right)−{a}\left(\mathrm{1}\right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{3}{k}−\mathrm{1}\right) \\ $$$${a}\left({n}+\mathrm{1}\right)−\mathrm{8}=\mathrm{3}×\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}−{n} \\ $$$${a}\left({n}+\mathrm{1}\right)=\frac{{n}\left(\mathrm{3}{n}+\mathrm{1}\right)}{\mathrm{2}}+\mathrm{8} \\ $$$$\Rightarrow{a}\left({n}\right)=\frac{\left({n}−\mathrm{1}\right)\left(\mathrm{3}{n}−\mathrm{2}\right)}{\mathrm{2}}+\mathrm{8} \\ $$$$\Rightarrow{a}\left(\mathrm{22}\right)=\frac{\mathrm{21}×\mathrm{64}}{\mathrm{2}}+\mathrm{8}=\mathrm{680} \\ $$

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