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.
Question 8(i)
Find the sum of nn terms of the series whose nthnth term is n3+3n.n3+3n.
Solution
The nhnh term is: an=n3+3nan=n3+3n Taking summation of the both sides n∑r=1ar=n∑r=1r3+n∑r=13r=[n(n+1)2]2+3(3n−1)3−1=n2(n+1)24+32(3n−1) Thus the sum of n terms is: Sn=n2(n+1)24+32(3n−1)
Question 8(ii)
Find the sum of n terms of the series whose nth term is 2n2+3n
Solution
The nth term is: an=2n2+3n Taking summation on the both sides n∑r=1ar=2n∑r=1r2+3n∑r=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
Question 8(iii)
Find the sum of n terms of the series whose nth term is n(n+1)(n+4)
Solution
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 n∑r=1ar=n∑r=1r3+5n∑r=1r2+4n∑r=1r=n2(n+1)24+5⋅n(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
Question 8(iv)
Find the sum of n terms of the series whose nth term is (2n−1)2
Solution
The nth term is: an=(2n−1)2an=4n2−4n+1 Taking summation of the both sides n∑r=1ar=4n∑r=1r2−4n∑r=1r+n∑r=11=4n(n+1)(2n+1)6−4n(n+1)2+n=n(n+1)2⋅[4(2n+1)6−2]+n=n(n+1)2⋅[8n+4−126]+n=n(n+1)(8n−8)12+n=4n(n+1)(2n−2)12+n=4(n(2n2−2n+2n−2))12+n=n(2n2−2)3+n=n(2n2−2+3)3=n(2n2+1)3 Thus the sum to n terms is: Sn=n(2n2+1)3
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