Find the sum of 100 terms of the series 1(3)+3(5)+5(7)+...
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31
Answer:
1353300
Step-by-step explanation:
Find the sum of 100 terms of the series 1(3)+3(5)+5(7)+...
1(3) + 3(5) + 5(7) +........................................+199(201)
= (2 -1)(2 + 1) + (4-1)(4 + 1) + (6-1)(6+1) +.......................................+(200-1)(200+1)
= (2² - 1) + (4² - 1) + (6²-1) +.................................................+ (200² - 1)
= (2² + 4² + 6² + ..................+200²) - 100
= 2²(1² + 2² + 3² +................+ 100²) - 100
sum of square of first n numbers = n(n+1)(2n+1)/6
= 4 * 100 * 101 * 201 /6 - 100
= 1353400 - 100
= 1353300
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