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Q.

The limn→∞ Sn if Sn=12n+14n2−1+14n2−4+…+13n2+2n−1 is

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a

π/2

b

2

c

1

d

π/6

answer is D.

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Detailed Solution

limn→∞ Sn=limn→∞ 1n14−0+14−1/n2+14−4/n2+…+4−n−1n2=limn→∞ 1n∑r=0n−1 14−(r/n)2=∫01 dx4−x2=sin−1⁡ x201=π6.
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The limn→∞ Sn if Sn=12n+14n2−1+14n2−4+…+13n2+2n−1 is