JEE Mathematics 2026 Question with Solution

Given: The coefficient of $x$ in $(ax^2+bx+c)(1-2x)^{26}$ is $-56$, and the coefficients of $x^2$ and $x^3$ are both zero.

Find: The value of $a+b+c$.

Use the first few terms of the binomial expansion:

So,

Now multiply by $(ax^2+bx+c)$ and compare coefficients.

Coefficient of $x$:

Coefficient of $x^2$:

Coefficient of $x^3$:

Dividing by $-52$,

Subtract the third equation from the second:

Substitute into $b-52c=-56$:

Hence,

Now use $a-25b+400c=0$:

Therefore,

The correct option is C.

Given: Only the coefficients of $x$, $x^2$, and $x^3$ matter.

Find: $a+b+c$.

Instead of expanding everything, keep only terms up to $x^3$ in $(1-2x)^{26}$:

This works because higher powers cannot contribute to the coefficients being compared.

Now directly write the three coefficient equations:

Subtract the last two equations:

Then from $b-52c=-56$, we get $c=3$ and hence $b=100$. Substituting back gives $a=1300$.

Therefore, $a+b+c=1403$, so the correct option is C.