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 \(j_4\)?

Author: Muft Shiksha Editorial Team Published:
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Correct Answer

B. इकतालीस(41)

Step 1

Concept

\(j_4=3+6+12+20=41\). In exams, simplify \(n^2+n\) first.

Step 2

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.

Step 3

Exam Tip

\(j_4=3+6+12+20=41\) है। परीक्षा में \(n^2+n\) को पहले सरल करें।

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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 \(j_4\)?

Correct Answer: B. इकतालीस / (41). Explanation: \(j_4=3+6+12+20=41\) है। परीक्षा में \(n^2+n\) को पहले सरल करें। / \(j_4=3+6+12+20=41\). In exams, simplify \(n^2+n\) first.

Which concept should I revise for this Mathematics MCQ?

\(j_4=3+6+12+20=41\). In exams, simplify \(n^2+n\) first.

What exam hint can help solve this Mathematics question?

\(j_4=3+6+12+20=41\) है। परीक्षा में \(n^2+n\) को पहले सरल करें।