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Question Number 89768 by john santu last updated on 19/Apr/20

t(n) − t(n−1) = n    for n > 0 and t(0) = 1   find t(n)

$${t}\left({n}\right)\:−\:{t}\left({n}−\mathrm{1}\right)\:=\:{n}\:\: \\ $$ $${for}\:{n}\:>\:\mathrm{0}\:{and}\:{t}\left(\mathrm{0}\right)\:=\:\mathrm{1}\: \\ $$ $${find}\:{t}\left({n}\right)\: \\ $$

Answered by john santu last updated on 19/Apr/20

Answered by mr W last updated on 19/Apr/20

t(k)−t(k−1)=k  Σ_(k=1) ^n t(k)−Σ_(k=1) ^n t(k−1)=Σ_(k=1) ^n k  Σ_(k=1) ^n t(k)−Σ_(k=0) ^(n−1) t(k)=Σ_(k=1) ^n k  Σ_(k=0) ^(n−1) t(k)+t(n)−t(0)−Σ_(k=0) ^(n−1) t(k)=Σ_(k=1) ^n k  t(n)−t(0)=Σ_(k=1) ^n k=((n(n+1))/2)  t(n)=t(0)+((n(n+1))/2)  ⇒t(n)=1+((n(n+1))/2)

$${t}\left({k}\right)−{t}\left({k}−\mathrm{1}\right)={k} \\ $$ $$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{t}\left({k}\right)−\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{t}\left({k}−\mathrm{1}\right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k} \\ $$ $$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{t}\left({k}\right)−\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{t}\left({k}\right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k} \\ $$ $$\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{t}\left({k}\right)+{t}\left({n}\right)−{t}\left(\mathrm{0}\right)−\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{t}\left({k}\right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k} \\ $$ $${t}\left({n}\right)−{t}\left(\mathrm{0}\right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$ $${t}\left({n}\right)={t}\left(\mathrm{0}\right)+\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$ $$\Rightarrow{t}\left({n}\right)=\mathrm{1}+\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$

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