Class 11MathematicsSequences and Seriesnth term of special series, Sum to n terms and Infinite number of terms
Difficultयदि $\left(\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots+\frac{1}{\alpha+1012}\right) - \left(\frac{1}{2 \cdot 1}+\frac{1}{4 \cdot 3}+\frac{1}{6 \cdot 5}+\ldots+\frac{1}{2024 \cdot 2023}\right) = \frac{1}{2024}$,तो $\alpha$ का मान ज्ञात कीजिए।
- A$1367$
- B$1058$
- C$1056$
- D$1011$
Explore More
Similar Questions
यदि किसी अनुक्रम का $n$ वाँ पद $T_n = 2n - 1$ है,तो $n$ पदों का योग $S_n = \dots$
Easy
View Solutionनिम्नलिखित श्रेणी $\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \dots$ का अनंत तक का योग क्या होगा?
Easy
View Solutionमाना $u_n = \frac{1}{\sqrt{5}} \left[ \left( \frac{1 + \sqrt{5}}{2} \right)^n - \left( \frac{1 - \sqrt{5}}{2} \right)^n \right]$,$n = 0, 1, 2, ...$ के लिए,तो निम्नलिखित में से कौन सा सत्य है?
Difficult
View Solutionयदि $\sum_{r=1}^{10} r! (r^3 + 6r^2 + 2r + 5) = \alpha(11!)$ है,तो $\alpha$ का मान ...... है.
DifficultJEE MAIN 2021
View Solution$\sum_{k=1}^n k(k+1)(k+2) \ldots(k+r-1) =$
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo