JEE Mains · Maths · STD 11 - 8. sequence and series
The sum \(1+3+11+25+45+71+.\). upto 20 terms, is equal to
- A \(7240\)
- B \(7130\)
- C \(6982\)
- D \(8124\)
Answer & Solution
Correct Answer
(A) \(7240\)
Step-by-step Solution
Detailed explanation
Given sum is \(S_n=1+3+11+25+45+71+\ldots+T_n\) First order differences are in A.P. Thus, we can assume that \(\mathrm{T}_{\mathrm{n}}=\mathrm{an}^2+\mathrm{bn}+\mathrm{c}\) Solving \(\left\{\begin{array}{c}T_1=1=a+b+c \\ T_2=3=4 a+2 b+c \\ T_3=11=9 a+3 b+c\end{array}\right\}\),…
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