JEE Mathematics 2026 Question with Solution

Given: $x^2+x+1=0$

Find: $\sum_{n=1}^{25}\left(x^n+\frac{1}{x^n}\right)^2$

From $x^2+x+1=0$, the roots are the complex cube roots of unity, $\omega$ and $\omega^2$. Take $x=\omega$. Then

using $\frac{1}{\omega}=\omega^2$.

Now use the periodic properties of cube roots of unity:

If $n$ is a multiple of $3$, say $n=3k$, then

If $n$ is not a multiple of $3$, then

so

From $1$ to $25$, the multiples of $3$ are $3,6,9,12,15,18,21,24$, so there are $8$ such terms. Hence the remaining $25-8=17$ terms are non-multiples of $3$.

Therefore,

Therefore, the working in the solution gives the value $49$. However, $49$ is not present among the given options A, B, C, D. The solution still marks the correct option as C, so there is a discrepancy between the computed result and the listed options. Using the solution's marked option, the answer is C.

Given: $x^2+x+1=0$

Find: the sum of squares up to $n=25$.

Since $x$ is a non-real cube root of unity, the expression

repeats with period $3$.

The pattern is:

with value $2$ when $n\equiv 0 \pmod{3}$ and value $-1$ otherwise. Squaring gives the repeating pattern

Between $1$ and $25$, there are $8$ multiples of $3$ and $17$ other integers. Hence

So the computed value is $49$. This does not match any option, which indicates an error in the provided options or answer key. The solution's nevertheless labels C as correct.