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Question Number 52079 by jojosehrawat21@gmail.com last updated on 03/Jan/19

Sum to the n terms of the series whose n^(th ) term is 2^(n−1 ) + 8n^3 −6n^2

Sumtothentermsoftheserieswhosenthtermis2n−1+8n3−6n2

Commented by maxmathsup by imad last updated on 03/Jan/19

let U_n =2^(n−1) +8n^3 −6n^2 with U_1 =3 ⇒ Σ_(k=1) ^n U_k =Σ_(k=1) ^n 2^(k−1) +8Σ_(k=1) ^n k^3 −6 Σ_(k=1) ^n k^2 =Σ_(k=0) ^(n−1) 2^k +8((n^2 (n+1)^2 )/4) −6 ((n(n+1)(2n+1))/6) =((1−2^n )/(1−2)) +2n^2 (n+1)^2 −n(n+1)(2n+1) =2^n −1 +n(n+1){2n(n+1)−2n−1} =2^n −1 +(n^2 +n){2n^2 −1} =2^n −1 +2n^4 −n^2 +2n^3 −n Σ_(k=1) ^n U_k =2^n +2n^4 +2n^3 −n^2 −n−1 .

letUn=2n−1+8n3−6n2withU1=3⇒∑k=1nUk=∑k=1n2k−1+8∑k=1nk3−6∑k=1nk2=∑k=0n−12k+8n2(n+1)24−6n(n+1)(2n+1)6=1−2n1−2+2n2(n+1)2−n(n+1)(2n+1)=2n−1+n(n+1){2n(n+1)−2n−1}=2n−1+(n2+n){2n2−1}=2n−1+2n4−n2+2n3−n∑k=1nUk=2n+2n4+2n3−n2−n−1.

Answered by tanmay.chaudhury50@gmail.com last updated on 03/Jan/19

T_n =2^(n−1) +8n^3 −6n^2 S=S_1 +S_2 −S_3 S_1 =((a(r^n −1))/(r−1))=((1(2^n −1))/(2−1))=2^n −1 S_2 =8×[((n(n+1))/2)]^2 =2n^2 (n+1)^2 S_3 =6×((n(n+1)(2n+1))/6)=n(n+1)(2n+1) S=2^(n−1) +n(n+1)(2n^2 +2n−2n−1) S=2^(n−1) +n(n+1)(2n^2 −1)

Tn=2n−1+8n3−6n2S=S1+S2−S3S1=a(rn−1)r−1=1(2n−1)2−1=2n−1S2=8×[n(n+1)2]2=2n2(n+1)2S3=6×n(n+1)(2n+1)6=n(n+1)(2n+1)S=2n−1+n(n+1)(2n2+2n−2n−1)S=2n−1+n(n+1)(2n2−1)

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