Question
Download Solution PDF100 या 100 से कम हरों वाली सभी उचित भिन्नों का योगफल होता है :
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFप्रयुक्त सूत्र :
(1 + 2 + 3 + 4 + ......... + 99) = [ n ( n + 1 ) ] ÷ 2
हल :
हम उन सभी उचित भिन्नों को लिख सकते हैं जिनके हर 100 या 100 से कम हैं:
⇒ 1/2 + (1/3 + 2/3) + (1/4 + 2/4 + 3/4) + ... (1/100 + 2/100 + .. .. 99/100)
⇒ 1/2 + 3/3 + 6/4 + ..... + 4950/100
⇒ 1/2 + 1 + 3/2 + 2 + ........ + 99/2
इस समीकरण को हम निम्न प्रकार से भी लिख सकते हैं:
⇒ 1/2 + 2/2 + 3/2 + ...... + 99/2
⇒ 1/2 ( 1 + 2 + 3 + ..... + 99 )
⇒ /2 [ ( 99 × 100) ÷ 2 ]
⇒ 99 × 25
⇒ 2475
अतः, सही उत्तर "2475" है।
Last updated on Jul 10, 2025
->HSSC CET Exam Date 2025 is 26th and 27th July 2025.
->The Haryana HSSC CET 2025 Exam will be held for two days in 4 shifts.
->Earlier, Haryana CET Group C Notice for EWS Certificate was out. A valid format of EWS Certificate has been given in the Notice.
-> Haryana CET Group C Notification 2025 was out on 26th May 2025.
-> The minimum educational qualification to apply for the Common Eligibility Test is 10+2/equivalent
-> Candidate applying for CET should not be less than 18 years of age and not more than 42 years.
-> Aspirants must go through the Haryana CET Previous Years’ Papers to understand the need for the exam and prepare for the exam in the right direction.