Question Number 198243 by mr W last updated on 15/Oct/23
$${find}\:{the}\:{sum}\:{of}\:{the}\:{first}\:{n}\:{terms}\:{from} \\ $$$$\mathrm{1},\:\mathrm{2}+\mathrm{3},\:\mathrm{4}+\mathrm{5}+\mathrm{6},\:\mathrm{7}+\mathrm{8}+\mathrm{9}+\mathrm{10},\:… \\ $$
Answered by som(math1967) last updated on 15/Oct/23
$$\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+\mathrm{5}+\mathrm{6}+\mathrm{7}+\mathrm{8}+\mathrm{9}+\mathrm{10}+..{n} \\ $$$${no}\:{of}\:{term}\:=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$${sum}\:{of}\:\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\:{terms}\: \\ $$$$=\frac{\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}×\left\{\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\:+\mathrm{1}\right\}}{\mathrm{2}} \\ $$$$=\frac{{n}\left({n}+\mathrm{1}\right)\left({n}^{\mathrm{2}} +{n}+\mathrm{2}\right)}{\mathrm{8}} \\ $$
Answered by universe last updated on 15/Oct/23
![1, 2+3, 4+5+6, 7+8+9+10, ... = {a_n } ((n^3 +n)/2) →this is a sum of n^(th) term of sequence {a_n } now P (let) = Σ_(n=1) ^n ((n^3 /2)+(n/2)) P = (1/2)×[((n(n+1))/2)]^2 +(1/2)×((n(n+1))/2) P = ((n(n+1)(n^2 +n+2))/8)](https://www.tinkutara.com/question/Q198248.png)
$$\mathrm{1},\:\mathrm{2}+\mathrm{3},\:\mathrm{4}+\mathrm{5}+\mathrm{6},\:\mathrm{7}+\mathrm{8}+\mathrm{9}+\mathrm{10},\:…\:\:\:=\:\:\:\left\{{a}_{{n}} \right\} \\ $$$$\:\:\:\:\frac{{n}^{\mathrm{3}} +{n}}{\mathrm{2}}\:\rightarrow{this}\:{is}\:{a}\:{sum}\:{of}\:{n}^{{th}} \:{term}\:{of}\: \\ $$$$\:\:{sequence}\:\left\{{a}_{{n}} \right\} \\ $$$$\:\:\:\:{now}\: \\ $$$$\:\:\:\:\:{P}\:\left({let}\right)\:=\:\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{{n}^{\mathrm{3}} }{\mathrm{2}}+\frac{{n}}{\mathrm{2}}\right)\:\: \\ $$$$\:\:\:\:{P}\:\:\:\:\:=\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}×\left[\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\right]^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}×\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\:\:\:{P}\:\:=\:\:\:\frac{{n}\left({n}+\mathrm{1}\right)\left({n}^{\mathrm{2}} +{n}+\mathrm{2}\right)}{\mathrm{8}}\: \\ $$$$\:\:\: \\ $$
Answered by mr W last updated on 15/Oct/23
![the n^(th) term has k_n =n numbers Σ_(i=1) ^(n−1) k_i =((n(n−1))/2) a_n =[((n(n−1))/2)+1]+[((n(n−1))/2)+2]+...+[((n(n−1))/2)+n] a_n =n×((n(n−1))/2)+1+2+...+n a_n =n×((n(n−1))/2)+((n(n+1))/2)=((n(n^2 +1))/2) s_n =Σ_(k=1) ^n a_k =Σ_(k=1) ^n ((k(k^2 +1))/2)=(1/2)Σ_(k=1) ^n (k+k^3 ) =(1/2)[((n(n+1))/2)+((n^2 (n+1)^2 )/4)] =((n(n+1)(n^2 +n+2))/8) ✓](https://www.tinkutara.com/question/Q198251.png)
$${the}\:{n}^{{th}} \:{term}\:{has}\:{k}_{{n}} ={n}\:{numbers} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}{k}_{{i}} =\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}} \\ $$$${a}_{{n}} =\left[\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}}+\mathrm{1}\right]+\left[\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}}+\mathrm{2}\right]+…+\left[\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}}+{n}\right] \\ $$$${a}_{{n}} ={n}×\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}}+\mathrm{1}+\mathrm{2}+…+{n} \\ $$$${a}_{{n}} ={n}×\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}}+\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}=\frac{{n}\left({n}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{2}} \\ $$$${s}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{k}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{k}\left({k}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left({k}+{k}^{\mathrm{3}} \right) \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}+\frac{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}\right] \\ $$$$\:\:\:=\frac{{n}\left({n}+\mathrm{1}\right)\left({n}^{\mathrm{2}} +{n}+\mathrm{2}\right)}{\mathrm{8}}\:\checkmark \\ $$