If 1² · ( ¹⁵ C₁ ) + 2² · ( ¹⁵ C₂ ) + 3² · ( ¹⁵ C₃ ) + + 15² · ( ¹⁵ C₁₅ ) = 2m · 3ⁿ · 5k$1^{2} \cdot \left( ^{15} \mathrm{C}_{1} \right) + 2^{2} \cdot \left( ^{15} \mathrm{C}_{2} \right) + 3^{2} \cdot \left( ^{15} \mathrm{C}_{3} \right) + \dots + 15^{2} \cdot \left( ^{15} \mathrm{C}_{15} \right) = 2^{\mathrm{m}} \cdot 3^{\mathrm{n}} \cdot 5^{\mathrm{k}}$, where m, n, k ∈ N, then m + n + k is equal to :-

Core Logic

To find the sum of all coefficients in a polynomial expansion, substitute x = 1$x = 1$.

Step 1: Evaluate Limit for b

Evaluate b = x → 0 ∫₀x ln(1 + t)1 + t²⁰²⁴ dtx²$b = \lim_{x \to 0} \frac{\int_{0}^{x} \frac{\ln(1 + t)}{1 + t^{2024}} dt}{x^2}$ Using L'Hôpital's Rule (differentiating numerator via Newton-Leibniz):

Step 2: Analyze Common Roots

The given second equation is 2bx² + ax + 4 = 0$2bx^2 + ax + 4 = 0$. Substitute a = 1$a = 1$ and b = (1)/(2)$b = \frac{1}{2}$:

The discriminant of x² + x + 4 = 0$x^2 + x + 4 = 0$ is D = 1 - 16 < 0$D = 1 - 16 < 0$. Roots are non-real complex conjugates.

Step 3: Final Ratio

Since c, d, e in R$c, d, e \in \mathbb{R}$ and one root is common with a quadratic having non-real roots, both roots must be common. Thus, the coefficients must be proportional:

This implies d : c : e = 1 : 1 : 4$d : c : e = 1 : 1 : 4$.

Pattern Recognition

If a quadratic equation with real coefficients shares a common root with another quadratic having complex roots (D < 0$D < 0$), both roots must be shared, meaning their coefficients are directly proportional.

Chapter Mix

Class 11 Maths: Binomial Theorem Class 12 Maths: Limits and Derivatives Class 11 Maths: Quadratic Equations

Core Logic

Step 1: Find Coefficient of x^3

To find coeff of x³$x^3$ in (1+x)¹⁷ - x(1+x)¹⁷x¹⁵$\frac{(1+x)^{17} - x(1+x)^{17}}{x^{15}}$, we need the coeff of x¹⁸$x^{18}$ in the numerator (1+x)¹⁷ - x(1+x)¹⁷$(1+x)^{17} - x(1+x)^{17}$. The maximum power of x$x$ in (1+x)¹⁷$(1+x)^{17}$ is 17$17$, and in x(1+x)¹⁷$x(1+x)^{17}$ is 18$18$. Coeff of x¹⁸$x^{18}$ in (1+x)¹⁷$(1+x)^{17}$ is 0. Coeff of x¹⁸$x^{18}$ in x(1+x)¹⁷$x(1+x)^{17}$ is the coeff of x¹⁷$x^{17}$ in (1+x)¹⁷$(1+x)^{17}$, which is 1717 = 1$\binom{17}{17} = 1$. Thus, coeff of x¹⁸$x^{18}$ in the numerator is 0 - 1 = -1$0 - 1 = -1$.

Step 2: Find Coefficient of x^{-13}

To find coeff of x⁻¹³$x^{-13}$, we need the coeff of x²$x^2$ in the numerator (1+x)¹⁷ - x(1+x)¹⁷$(1+x)^{17} - x(1+x)^{17}$. Coeff of x²$x^2$ in (1+x)¹⁷$(1+x)^{17}$ is 172$\binom{17}{2}$. Coeff of x²$x^2$ in x(1+x)¹⁷$x(1+x)^{17}$ is coeff of x¹$x^1$ in (1+x)¹⁷$(1+x)^{17}$, which is 171$\binom{17}{1}$. Value = 172 - 171 = (17 × 16)/(2) - 17 = 136 - 17 = 119$\binom{17}{2} - \binom{17}{1} = \frac{17 \times 16}{2} - 17 = 136 - 17 = 119$.

Step 3: Final Sum

Sum of coefficients = -1 + 119 = 118$-1 + 119 = 118$.

Chapter Mix

Class 11 Maths: Binomial Theorem