If 2 2 -5 =1 has exactly 7 solutions in the interval [0, n 2 ], for the least value of n , then k=1n k2k is equal to:
- 1214 (215-15 )
- 1- 15213
- 1215 (214-14 )
- 1213 (214-15 )
Answer
(4) 1213 (214-15 )
Solution
2 ( 2 -1 )-5 -1=0 (2 +1)( -3)=0
∵ | | 1: =3, i.e. = 13
[figure]
7th solution is 6 + -1 13 (between 6 and 13 2) least n=13
S= k=113 k2k: S- S2= ( 12+ 122+ + 1213 )- 13214
S2=1- 1213- 13214
S=2- 15213= 1213 (214-15 )
Answer: (4)