Test Index

Sequences and Series

Section: Mathematics

© examsnet.com

Question : 7 of 25

Marks: +1, -0

If

2.5+5.9+8.13+11.17+\dots

= an^{3}+bn^{2}+cn+d

a-b+c-d=

[AP EAPCET 21 May 2025 S2]

7
5
-3
-1

Solution:

Given,

2 \cdot 5+5 \cdot 9+8 \cdot 13+11 \cdot 17+\dots+

upto

n

terms

\;t_n=(3n-1)(4n+1)

\;t_n=12n^{2}-n-1

\;\therefore S_n=\sum t_n=\sum (12n^{2}-n-1)=12\sum n^{2}-\sum n-\sum 1

\;=12 \cdot \frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2}-n

\;=n\left[2(2n^{2}+3n+1)-\frac{n+1}{2}-1\right]=\frac{n}{2}\left[8n^{2}+12n+4-n-1-2\right]

\;=\frac{n}{2}[8n^{2}+11n+1]

\;=4n^{3}+\frac{11}{2}n^{2}+\frac{n}{2}=an^{3}+bn^{2}+cn+d

\;\Rightarrow a=4, b=\frac{11}{2}, c=\frac{1}{2}\;\text{ and }\;d=0

\;\therefore a-b+c-d=4-\frac{11}{2}+\frac{1}{2}-0=4-5=-1

© examsnet.com

Go to Question:

Prev Question

Next Question