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The value of limn1nsinπn+sin2πn++sinnπn is

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a
π
b
-π2
c
d

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detailed solution

Correct option is D

Read. limit=limn→∞ 1n∑n=1n sin⁡rπn            =∫01 sin⁡πxdx=−cos⁡πxπ01=−1π[−1−1]=2π


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limn1n+1+1n+2++16n is equal to


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