Question
=limn→∞[1n+1√n2+n+1√n2+2n+⋯+1√n2+(n−1)n] is equal to [RPET 2000]
Answer: Option B
:
B
y=limn→∞[1n+1√n2+n+⋯+1√n2+(n−1)n]
⇒y=limn→∞⎡⎢⎣1n+1n√1+1n+⋯+1n√1+(n−1)n⎤⎥⎦
⇒y=1nlimn→∞⎡⎢⎣1+1√1+1n+⋯+1√1+(n−1)n⎤⎥⎦
y=limn→∞1n∑k=1n1√1+(k−1)n,Putk−1n=xand1n=dx
⇒y=limn→∞∫n−1n0dx√1+x=limn→∞2[√1+x](n−1n)0
⇒y=2limn→∞[√2n−1n−1]=2limn→∞√2n−1n−2
⇒y=2limn→∞√2−1n−2=2√2−2
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:
B
y=limn→∞[1n+1√n2+n+⋯+1√n2+(n−1)n]
⇒y=limn→∞⎡⎢⎣1n+1n√1+1n+⋯+1n√1+(n−1)n⎤⎥⎦
⇒y=1nlimn→∞⎡⎢⎣1+1√1+1n+⋯+1√1+(n−1)n⎤⎥⎦
y=limn→∞1n∑k=1n1√1+(k−1)n,Putk−1n=xand1n=dx
⇒y=limn→∞∫n−1n0dx√1+x=limn→∞2[√1+x](n−1n)0
⇒y=2limn→∞[√2n−1n−1]=2limn→∞√2n−1n−2
⇒y=2limn→∞√2−1n−2=2√2−2
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