किसी अनुक्रम में \(j_1=3\) और \(j_n=j_{n-1}+n^2+n\) है। \(j_4\) का मान क्या होगा?
In a sequence, \(j_1=3\) and \(j_n=j_{n-1}+n^2+n\). What is the value of \(j_4\)?
Explanation opens after your attempt
B. इकतालीस(41)
Concept
\(j_4=3+6+12+20=41\). In exams, simplify \(n^2+n\) first.
Why this answer is correct
The correct answer is B. इकतालीस / (41). \(j_4=3+6+12+20=41\). In exams, simplify \(n^2+n\) first.
Exam Tip
\(j_4=3+6+12+20=41\) है। परीक्षा में \(n^2+n\) को पहले सरल करें।
Login to save your score, XP, coins and progress.
