0 reads0 ratings0 helpful
If (a_n=2n^2-3n+4), what is (a_{n+1}-a_n)?यदि (a_n=2n^2-3n+4), तो (a_{n+1}-a_n) क्या होगा?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: \(4n-1\)\(4n-1\)
Given \(a_n=2n^2-3n+4\), replace \(n\) by \(n+1\): \(a_{n+1}=2(n+1)^2-3(n+1)+4=2n^2+n+3\). Therefore, \(a_{n+1}-a_n=(2n^2+n+3)-(2n^2-3n+4)=4n-1\). The option \(4n+1\) can result from an error while subtracting the constant terms. Exam tip: write \(a_{n+1}\) separately before subtracting \(a_n\).दिया है \(a_n=2n^2-3n+4\)। \(n\) के स्थान पर \(n+1\) रखने पर \(a_{n+1}=2(n+1)^2-3(n+1)+4=2n^2+n+3\) मिलता है। अतः \(a_{n+1}-a_n=(2n^2+n+3)-(2n^2-3n+4)=4n-1\)। विकल्प \(4n+1\) में स्थिर पदों का घटाव गलत होने की संभावना रहती है। परीक्षा टिप: पहले \(a_{n+1}\) अलग से लिखकर फिर \(a_n\) घटाएँ।
Student feedback
Was this question useful?
Was this question useful?
👍 0 Helpful 👎 0 Not helpful
Difficulty
Easy0%
Medium0%
Hard0%
Was the explanation clear?
Yes 0% No 0%
0 responsesStudent Reviews
No published reviews yet.
