If the coefficient of x in the expansion of (ax² + bx + c)(1 − 2x)²⁶ is -56 and the coefficients of x² and x³ are both zero, then a + b + c is equal to

For any positive integer $n$, the expansion of $(1+x)^n$ is: $$(1+x)^n = \sum_{r=0}^{n}\binom{n}{r}x^r$$ General term: $T_{r+1} = \binom{n}{r}a^{n-r}b^r$ in $(a+b)^n$

$T_{r+1} = \binom{n}{r}(-2)^r x^r$ for $(1-2x)^n$

When finding coeff of $x^k$ in $(ax^2+bx+c)\cdot Q(x)$, consider all term-pairs whose degrees add to $k$:

Coeff of $x^1$ = $c\cdot[\text{coeff of }x^1\text{ in }Q] + b\cdot[\text{coeff of }x^0\text{ in }Q]$
Coeff of $x^2$ = $c\cdot[x^2] + b\cdot[x^1] + a\cdot[x^0]$
Coeff of $x^3$ = $c\cdot[x^3] + b\cdot[x^2] + a\cdot[x^1]$

This selective multiplication avoids full expansion — a key JEE time-saver.

$\binom{26}{1} = 26$ $\binom{26}{2} = 325$ $\binom{26}{3} = 2600$

Remember: $\binom{n}{2} = \frac{n(n-1)}{2}$, $\binom{n}{3} = \frac{n(n-1)(n-2)}{6}$

$P(x) = ax^2+bx+c$ has 3 unknowns. We need exactly 3 independent equations. The problem gives:
• Coeff of $x = -56$ → Eq (1)
• Coeff of $x^2 = 0$ → Eq (2)
• Coeff of $x^3 = 0$ → Eq (3)

3 independent linear equations in 3 unknowns always yield a unique solution (by linear algebra).

❌ Mistake 1: Forgetting $(-2)^r$ sign — always apply the sign alternation.

❌ Mistake 2: Missing cross-product terms — all combinations of factors whose degrees add to the target power must be included.

❌ Mistake 3: Arithmetic error in $\binom{26}{2}$ — it’s $325$, not $26^2 = 676$.

❌ Mistake 4: Not verifying $c=3$ in Eq(1) before proceeding to find $a$.

✅ Trick 1 (Integer constraint): Once you get $b = \frac{100c}{3}$, immediately try smallest integer $c$ divisible to give integer $b$. Here $c=3$ works instantly.

✅ Trick 2 (Eliminate smartly): Subtract equations sharing the same variable with coefficient 1 to eliminate it first. This reduces the system faster.

✅ Trick 3 (Answer check): $a=1300=\binom{26}{2}\times4$, which is the coeff of $x^2$ in $(1-2x)^{26}$. Recognising such patterns confirms correctness instantly.

Binomial Theorem is a guaranteed topic in every JEE Main session — typically 1–2 questions. This problem-type (polynomial × binomial expansion with zero-coefficient conditions) is a high-frequency pattern. It tests:
• Binomial general term recall
• Algebraic system-solving speed
• Calculation accuracy under time pressure

In JEE Advanced, similar ideas appear in harder forms with summation of coefficients or finding the greatest term. Regular practice of this pattern ensures 3–4 guaranteed marks per session.

$(1-x)^n = \sum_{r=0}^{n}\binom{n}{r}(-1)^r x^r$ $T_{r+1}=\binom{n}{r}a^{n-r}b^r$ $\binom{n}{r}=\frac{n!}{r!(n-r)!}$ Pascal: $\binom{n}{r}+\binom{n}{r-1}=\binom{n+1}{r}$ Sum of coefficients: put $x=1$

The system of equations formed here is:
Eq1: $b – 52c = -56$
Eq2: $a – 52b + 1300c = 0$
Eq3: $a – 25b + 400c = 0$

This is a $3\times3$ linear system. The elimination method used (subtracting Eq3 from Eq2 to eliminate $a$) is identical to Gaussian elimination — a technique also relevant in Linear Algebra for JEE Advanced.

To master this type:
• Memorise coefficients of $(1+x)^n$ for small $r$ values (0 to 4)
• Practice 5 “coefficient extraction from product” problems daily for 1 week
• Solve at least 3 “zero-coefficient” system problems before the exam
• Time yourself: this type should be solved in under 3.5 minutes in exam