Question

add 2sqrt(5) - 3sqrt(2) + 4sqrt(5)​

Answer 1

Answer:

No. But it’s really close.

The left hand side may be written as

X=∏∞k=2k1/2k−1

Take its logarithm

lnX=∑∞k=2lnk2k−1

Performing the sum using Sum[Log[n]/2^(n-1),{n,2,Infinity}] - Wolfram|Alpha Results

lnX≈1.01567

If the equation was correct, then lnX would have been equal to 1.

Answer 2

Step-by-step explanation:

hope you all understand