Question 8 Review Exercise

Solutions of Question 8 of Review Exercise of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Find the sum of nn terms of the series whose nthnth term is n3+3n.n3+3n.

The nhnh term is: an=n3+3nan=n3+3n Taking summation of the both sides nr=1ar=nr=1r3+nr=13r=[n(n+1)2]2+3(3n1)31=n2(n+1)24+32(3n1) Thus the sum of n terms is: Sn=n2(n+1)24+32(3n1)

Find the sum of n terms of the series whose nth term is 2n2+3n

The nth term is: an=2n2+3n Taking summation on the both sides nr=1ar=2nr=1r2+3nr=1r=2[n(n+1)(2n+1)6]++3n(n+1)2=n(n+1)(2n+1)3+3n(n+1)2=n(n+1)(2n+1)3+32=n(n+1)2(2n+1)+96=n(n+1)(4n+11)6 Thus sum to n terms is: Sn=n(n+1)(4n+11)6

Find the sum of n terms of the series whose nth term is n(n+1)(n+4)

The nth  term is: an=n(n+1)(n+4)an==n(n2+5n+4)an=n3+5n2+4n Taking summation of the both sides nr=1ar=nr=1r3+5nr=1r2+4nr=1r=n2(n+1)24+5n(n+1)(2n+1)6+4n(n+1)2=n(n+1)2[n(n+1)2+10n+53+4]=n(n+1)2[3n(n+1)+2(10n+5)+246]=n(n+1)2[3n2+3n+20n+10+246]=n(n+1)(3n2+23n+34)12 Thus the sum to n terms is: Sn=n(n+1)(3n2+23n+34)12

Find the sum of n terms of the series whose nth term is (2n1)2

The nth term is: an=(2n1)2an=4n24n+1 Taking summation of the both sides nr=1ar=4nr=1r24nr=1r+nr=11=4n(n+1)(2n+1)64n(n+1)2+n=n(n+1)2[4(2n+1)62]+n=n(n+1)2[8n+4126]+n=n(n+1)(8n8)12+n=4n(n+1)(2n2)12+n=4(n(2n22n+2n2))12+n=n(2n22)3+n=n(2n22+3)3=n(2n2+1)3 Thus the sum to n terms is: Sn=n(2n2+1)3