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If 1 + (1/2) + (1/3) + … + (1/20) = k, then what is the value of (1/4) + (1/6) + (1/8) + … + (1/40)?
- Ak/2
- B2k
- C(k - 1)/2
- D(k + 1)/2
Answer not set
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0
CCommunity
(k-1)/2
0
CCommunity
(1/2)+(1/30)+.............+(1/20)=k-1
and this series becomes 1/(n+1).........=k-1 now this is eq1
now take the other given condition:
(1/4)+(1/6).....+(1/20) now this equals to 1/(2n+2) which we can writw that
1/2(n+1) which is 1/2 times of eq 1.
so the answer is (k-1)/2
and this series becomes 1/(n+1).........=k-1 now this is eq1
now take the other given condition:
(1/4)+(1/6).....+(1/20) now this equals to 1/(2n+2) which we can writw that
1/2(n+1) which is 1/2 times of eq 1.
so the answer is (k-1)/2