Question

xn=n^3-6n^3+13n-7 write the sequence​

Answer 1

Answer:

Hello friend

Step-by-step explanation:

xn+1-xn = {(n+1)3-6(n+1)2+13(n+1)-7}-{ n3 – 6n2+ 13n – 7}

xn+1-xn = {(n+1)3-6(n+1)2+13(n+1)-7}-{ n3 – 6n2+ 13n – 7}= (n+1)3-n3-6{(n+1)2-n2}+13(n+1-n)-7+7

xn+1-xn = {(n+1)3-6(n+1)2+13(n+1)-7}-{ n3 – 6n2+ 13n – 7}= (n+1)3-n3-6{(n+1)2-n2}+13(n+1-n)-7+7= (n+1-n)(n2+2n+1+n2+n+n2)-6(2n+1)+13

xn+1-xn = {(n+1)3-6(n+1)2+13(n+1)-7}-{ n3 – 6n2+ 13n – 7}= (n+1)3-n3-6{(n+1)2-n2}+13(n+1-n)-7+7= (n+1-n)(n2+2n+1+n2+n+n2)-6(2n+1)+13= 1× (3n2+3n+1)-12n+7

xn+1-xn = {(n+1)3-6(n+1)2+13(n+1)-7}-{ n3 – 6n2+ 13n – 7}= (n+1)3-n3-6{(n+1)2-n2}+13(n+1-n)-7+7= (n+1-n)(n2+2n+1+n2+n+n2)-6(2n+1)+13= 1× (3n2+3n+1)-12n+7= 3n2-9n+8

xn+1-xn = {(n+1)3-6(n+1)2+13(n+1)-7}-{ n3 – 6n2+ 13n – 7}= (n+1)3-n3-6{(n+1)2-n2}+13(n+1-n)-7+7= (n+1-n)(n2+2n+1+n2+n+n2)-6(2n+1)+13= 1× (3n2+3n+1)-12n+7= 3n2-9n+8Which is not independent of n. Hence, the sequence is not an AP.