Binomial Theorem — JEE Main practice

1. Expand $(ax^2+bx+c)(1-2x)^{26}$ term by term and extract the coefficient of $x$ from each piece. Given the general term of $(1-2x)^{26}$ is $\binom{26}{k}(-2)^kx^k$, the coefficient of $x$ (from $bx$ times the constant term and $c$ times the $x^1$ term) is

$$b-52c = -56$$

2. Similarly extract the coefficient of $x^2$, which comes from $a$ times the constant term, $b$ times the $x^1$ term, and $c$ times the $x^2$ term of $(1-2x)^{26}$. Given this coefficient is zero,

$$a-52b+1300c = 0$$

3. Similarly extract the coefficient of $x^3$, which involves the $x^0$, $x^1$, $x^2$ terms shifted appropriately. Given this coefficient is also zero,

$$-52a+1300b-20800c = 0 \quad\Rightarrow\quad -a+25b-400c=0$$

4. Solve this system of three equations simultaneously for $a$, $b$, $c$. Solving gives

$$a=1300, \quad b=100, \quad c=3$$

5. Sum these values. Hence,

$$a+b+c = 1300+100+3 = 1403$$

Hence, $a+b+c=1403$, so the answer is option B.

1. Let $S$ denote the given sum, and use the standard technique of multiplying by $(1+x)$ and subtracting to telescope the series. Given

$$S = (1+x)+2(1+x)^2+\cdots+100(1+x)^{100}$$ $$(1+x)S = (1+x)^2+\cdots+99(1+x)^{100}+100(1+x)^{101}$$

2. Subtract $S$ from $(1+x)S$, so most terms telescope, leaving a simpler geometric-like expression. Therefore,

$$-xS = (1+x)\left(\frac{(1+x)^{100}-1}{x}\right)-100(1+x)^{101}$$

3. Divide through by $-x$ and simplify to isolate $S$. Hence,

$$S = \frac{(1+x)-(1+x)^{101}}{x^2}+\frac{100(1+x)^{101}}{x}$$

4. Extract the coefficient of $x^{48}$ from this expression by examining each term's contribution. Hence,

$$\text{Coeff of }x^{48} = -\binom{101}{50}+100\binom{101}{49}$$

Hence, the coefficient of $x^{48}$ is $100\binom{101}{49}-\binom{101}{50}$, so the answer is option D.

1. Rewrite $P_n$ using the identity $\binom{n}{r}\frac{1}{r+1}=\frac{1}{n+1}\binom{n+1}{r+1}$ so the sum converts into a single binomial expansion in terms of $n+1$. Given

$$P_n = \sum_{r=0}^{n}\frac{\binom{n}{r}(-2)^r}{r+1} = \frac{1}{n+1}\sum_{r=0}^{n}\binom{n+1}{r+1}(-2)^r$$

2. Multiply and divide by $-2$ to align the sum with the binomial expansion of $(1-2)^{n+1}$. Therefore,

$$P_n = \frac{-1}{2(n+1)}\sum_{r=0}^{n}\binom{n+1}{r+1}(-2)^{r+1} = \frac{-1}{2(n+1)}\left[(1-2)^{n+1}-1\right]$$

3. Simplify using $(1-2)^{n+1}=(-1)^{n+1}$. Hence,

$$P_n = \frac{1}{2(n+1)}\left[1-(-1)^{n+1}\right]$$

4. Evaluate this at even arguments $n=2n$, since we need $P_{2n}$. Given $(-1)^{2n+1}=-1$,

$$P_{2n} = \frac{1}{2(2n+1)}[1-(-1)] = \frac{1}{2n+1}$$

5. Take the reciprocal and sum from $n=1$ to $25$, since we need $\sum\frac{1}{P_{2n}}$. Therefore,

$$\sum_{n=1}^{25}\frac{1}{P_{2n}} = \sum_{n=1}^{25}(2n+1) = 3+5+\cdots+51$$

6. Use the arithmetic series sum formula. Hence,

$$\frac{25}{2}(3+51) = 25\times27 = 675$$

Hence, the value of the sum is 675, so the answer is option D.