1oye yeh sum dekhkar daro mat .. solve karo .. kyunki darr ke aage jeet hai !!
find .... [ k = 1Σ80( 1/√k ) ] .. where [.] denotes greatest integer function ...
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11 Answers
11nahi complete thing hai ...
y = [ 1 + 1/√2 + 1/√3 + ................. ]
11tapan masti mat kar
its not that easy
[1+1/√2+1/√3.......................+1/√80]
and not wat u have written
1it is very easy
we see that
1/√k > 2(√(k+1) -√k)
as
it reduces to
1/√k > 2/(√k+1 + √k) => √k+1 >√k which is true ..
and
similarly
1/√k < 2(√k-√(k-1))
so 1/√1 + 1/√2 + ... 1/√80 > 2(√81 - 1)=16
and
1/√1 + 1/√2 +....1/√80 < 1 + 2(√80-1) = √320 -1 = 16.xyz
thus answer =16
11/√k < 2(√k - √(k-1))
implies
1/√k < 2 /(√k + √(k-1))
which on cross multiplying gives .
√k + √(k-1)<2√k
that is = √k > √(k-1)
21frm wer do v get this : "1/√k > 2(√(k+1) -√k) "
Is it simply by observation or by sum mathematical proof??