JEE Main20254 Apr 2025Evening ShiftMathematicsComplex NumberActual
If is a root of the equation x^2+x+1=0 and _ k =1 ^ n ( ^ k + 1 ^ k )^2=20 , then n is equal to
Correct answer
0
Step-by-step solution
aligned & = & ( ^k+ 1 ^k )^2= ^ 2 k + 1 ^ 2 k +2 & = ^ 2 k + ^k+2 ^ 3 k =1 aligned aligned & _ k =1 ^ n ( ^ 2 k + ^ k +2 )=20 & ( ^2+ ^4+ ^6+ + ^ 2 n )+ ( + ^2+ ^3+ + . & . ^ n )+2 n =20 aligned Now if n=3 m, m I Then 0+0+2 n =20 n =10 (not satisfy) if n=3 m+1 , then aligned & ^2+ +2 n =20 & -1+2 n =20 n = 21 2 ( not possible ) aligned if n=3 m+2 , aligned & ( ^8+ ¹⁰ )+ ( ^4+ ^5 )+2 n =20 & ( ^2+ )+ ( + ^2 )+2 n =20 & 2 n =22 & n =11 satisfy n =3 ~m +2 & n =11 aligned