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International Baccalaureatte | Further Maths HL
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how to do this
Date Posted: 3 years ago
Views: 405
lim n→∞ ∑(k=1, to n) ( k^8 / n^9 + √k / n√n )
= lim n→∞ ∑(k=1, to n) 1/n ( k^8 / n^8 + √k / √n )
= lim n→∞ 1/n ∑(k=1,to n) ( (k/n)^8 + √(k/n) )
Or lim n→∞ 1/n ∑(k=1,to n) ( (k(1/n))^8 + √(k(1/n)) )
= ∫(lower limit 0, upper limit 1) (k^8 + √k) dk
= [ 1/9 k^9 + ⅔ k³/² ] (upper limit 1, lower limit 0)
= (1/9 (1^9) + ⅔ (1³/²) ) - (1/9 (0^9) + ⅔ (0³/²) )
= 1/9 + ⅔
= 1/9 + 6/9
= 7/9
= lim n→∞ ∑(k=1, to n) 1/n ( k^8 / n^8 + √k / √n )
= lim n→∞ 1/n ∑(k=1,to n) ( (k/n)^8 + √(k/n) )
Or lim n→∞ 1/n ∑(k=1,to n) ( (k(1/n))^8 + √(k(1/n)) )
= ∫(lower limit 0, upper limit 1) (k^8 + √k) dk
= [ 1/9 k^9 + ⅔ k³/² ] (upper limit 1, lower limit 0)
= (1/9 (1^9) + ⅔ (1³/²) ) - (1/9 (0^9) + ⅔ (0³/²) )
= 1/9 + ⅔
= 1/9 + 6/9
= 7/9
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My solution
Date Posted:
3 years ago
thanks so much! i understand now
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