Solution:
2.25≤2+2n+2≤202
2.25−2≤2+2n+2−2≤202−2
0.25≤2n+2≤200
log20.25≤n+2≤log2200
−2≤n+2≤7.xx
−4≤n≤7.xx−2
−4≤n≤5.xx
Possible integers=−4,−3,−2,−1,0,1,2,3,4,5 If we see the second expression that is provided, i.e
3+3n+1,it can be implied that n should be at least −1 for this expression to be an integer.
So, n=−1,0,1,2,3,4,5
Hence, there are a total of 7 values.
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