We need to find the coefficients $a$, $b$, and $c$ from the expansion of the product $(ax^2 + bx + c)(1 - 2x)^{26}$. We first find the first few terms of the binomial expansion of $(1 - 2x)^{26}$ using the binomial theorem: $(1 + y)^n = \sum_{k=0}^{n} \binom{n}{k} y^k$.
Here, $y = -2x$ and $n = 26$. The expansion is:
$(1 - 2x)^{26} = \binom{26}{0} (1)^{26}(-2x)^0 + \binom{26}{1} (1)^{25}(-2x)^1 + \binom{26}{2} (1)^{24}(-2x)^2 + \binom{26}{3} (1)^{23}(-2x)^3 + \dots$
Calculating the terms:
So, the expansion starts as: $(1 - 2x)^{26} = 1 - 52x + 1300x^2 - 20800x^3 + \dots$
Now, consider the product $(ax^2 + bx + c)(1 - 52x + 1300x^2 - 20800x^3 + \dots)$. We extract the coefficients for $x$, $x^2$, and $x^3$:
We solve the system of three linear equations:
The coefficients are $a=1300$, $b=100$, and $c=3$.
We need to find the value of $a + b + c$.
$a + b + c = 1300 + 100 + 3 = 1403$The calculated value is 1403, which corresponds to Option 3. However, based on the provided answer, the correct option is B.
Let A = {-3, -2, -1, 0, 1, 2, 3}. Let R be a relation on A defined by xRy if and only if $0\le x^2+2y\le 4$. Let $l$ be the number of elements in R and $m$ be the minimum number of elements required to be added in R to make it a reflexive relation. Then $l + m$ is equal to
Let $\alpha$ and $\beta$ be the roots of $x^2 + \sqrt{3}x-16=0$, and $\gamma$ and $\delta$ be the roots of $x^2 + 3x - 1 = 0$. If $P_n = \alpha^n + \beta^n$ and $Q_n = \gamma^n + \delta^n$, then $\frac{P_{25} + \sqrt{3}P_{24}}{2P_{23}} + \frac{Q_{25}-Q_{23}}{Q_{24}}$ is equal to
Let the domain of the function $f (x) = \log_2 \log_4 \log_6(3 + 4x -x^2)$ be $(a, b)$.
If $\int_{b-a}^{b+a}[x^2] dx=p-\sqrt{q}-\sqrt{r}$, $p,q,r\in N$, $gcd(p,q,r) = 1$, where $[.]$ is the greatest integer function, then $p + q + r$ is equal to
Let A be a matrix of order $3 \times 3$ and $|A| = 5$. If $|2\text{adj} (3A \text{adj} (2A))| = 2^\alpha \cdot 3^\beta \cdot 5^\gamma$, $\alpha, \beta, \gamma \in N$, then $\alpha + \beta + \gamma$ is equal to
Let $a_1, a_2, a_3,....$ be a G.P. of increasing positive numbers. If $a_3a_5 = 729$ and $a_2 + a_4 = \frac{111}{4}$, then $24 (a_1 + a_2 + a_3)$ is equal to
All five letter words are made using all the letters A, B, C, D, E and arranged as in an English dictionary with serial numbers. Let the word at serial number $n$ be denoted by $W_n$. Let the probability $P(W_n)$ of choosing the word $W_n$ satisfy $P(W_n) = 2P(W_{n-1})$, $n > 1$.
If $P(CDBEA) = \frac{2^\alpha}{2^\beta-1}$, $\alpha, \beta\in N$, then $\alpha + \beta$ is equal to :
Let $a \in \mathbf{R}$ and $A$ be a matrix of order $3 \times 3$ such that $\det (A) = -4$ and $A + I = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix}$, where $I$ is the identity matrix of order $3 \times 3$. If $\det ((a+1)\text{adj}((a-1)A))$ is $2^m 3^n$, $m, n \in \{0, 1, 2, \dots, 20\}$, then $m+n$ is equal to :
Let A = {-3, -2, -1, 0, 1, 2, 3}. Let R be a relation on A defined by xRy if and only if $0\le x^2+2y\le 4$. Let $l$ be the number of elements in R and $m$ be the minimum number of elements required to be added in R to make it a reflexive relation. Then $l + m$ is equal to
Let $\alpha$ and $\beta$ be the roots of $x^2 + \sqrt{3}x-16=0$, and $\gamma$ and $\delta$ be the roots of $x^2 + 3x - 1 = 0$. If $P_n = \alpha^n + \beta^n$ and $Q_n = \gamma^n + \delta^n$, then $\frac{P_{25} + \sqrt{3}P_{24}}{2P_{23}} + \frac{Q_{25}-Q_{23}}{Q_{24}}$ is equal to
Let the domain of the function $f (x) = \log_2 \log_4 \log_6(3 + 4x -x^2)$ be $(a, b)$.
If $\int_{b-a}^{b+a}[x^2] dx=p-\sqrt{q}-\sqrt{r}$, $p,q,r\in N$, $gcd(p,q,r) = 1$, where $[.]$ is the greatest integer function, then $p + q + r$ is equal to
Let A be a matrix of order $3 \times 3$ and $|A| = 5$. If $|2\text{adj} (3A \text{adj} (2A))| = 2^\alpha \cdot 3^\beta \cdot 5^\gamma$, $\alpha, \beta, \gamma \in N$, then $\alpha + \beta + \gamma$ is equal to
Let $a_1, a_2, a_3,....$ be a G.P. of increasing positive numbers. If $a_3a_5 = 729$ and $a_2 + a_4 = \frac{111}{4}$, then $24 (a_1 + a_2 + a_3)$ is equal to