The coefficient of x and x\(^2\) in (1 + x)\(^p\) (1 – x)\(^q\) are 4 and – 5, then 2p + 3q is
Correct Answer: 63
Solution and Explanation
The correct answer is 63
\((1+x)^p(1-x)^q\)
\((1+px+\frac{p(p-1)}{2!}x^2+......)\)
\((1-qx+\frac{q(q-1)}{2!}x^2-......)\)
\(p-q=4\)
\(\frac{p(p-1)}{2}+\frac{q(q-1)}{2}-pq=-5\)
p2+q2 - p - 2 - 2pq=-10
(q+4)2 + q2 - (q-4) - q - 2(4+q)q = -10
q2 + 8q + 16 - q2 - q - 4 - q - 8q - 2q2 = -10
-2q=-22
so , q=11 and p=15
\(\therefore \) 2p + 3q = 2(15) + 3(11)
= 30 + 33
= 63
Top JEE Main Mathematics Questions
- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
- If , then the general value of is:
- The general solution of
- If the constant term in the binomial expansion of $\left(\frac{x^{\frac{3}{2}}}{2}-\frac{4}{x^l}\right)^9$ is $-84$ and the coefficient of $x^{-3 l}$ is $2^\alpha \beta$, where $\beta<0$ is an odd number, then $|\alpha l-\beta|$ is equal to_____
Top JEE Main Binomial theorem Questions
- The number of integral terms in the expansion of (31/2 + 51/4)680 is equal to:
- Suppose \( 2 - p \), \( p \), \( 2 - \alpha \), \( \alpha \) are the coefficients of four consecutive terms in the expansion of \( (1 + x)^n \). Then the value of \( p^2 - \alpha^2 + 6\alpha + 2p \) equals
- Let \( \alpha = \sum_{k=0}^{n} \left( \frac{\binom{n}{k}}{k+1} \right)^2 \) and \( \beta = \sum_{k=0}^{n-1} \left( \frac{\binom{n}{k} \binom{n}{k+1}}{k+2} \right) \).
If \( 5\alpha = 6\beta \), then \( n \) equals __________. - \(^{n-1}C_r = (k^2 - 8) ^{n}C_{r+1}\) if and only if:
- If \( A \) denotes the sum of all the coefficients in the expansion of \( (1 - 3x + 10x^2)^n \) and \( B \) denotes the sum of all the coefficients in the expansion of \( (1 + x^2)^n \), then:
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
Concepts Used:
Binomial Expansion Formula
The binomial expansion formula involves binomial coefficients which are of the form
(n/k)(or) nCk and it is calculated using the formula, nCk =n! / [(n - k)! k!]. The binomial expansion formula is also known as the binomial theorem. Here are the binomial expansion formulas.
This binomial expansion formula gives the expansion of (x + y)n where 'n' is a natural number. The expansion of (x + y)n has (n + 1) terms. This formula says:
We have (x + y)n = nC0 xn + nC1 xn-1 . y + nC2 xn-2 . y2 + … + nCn yn
General Term = Tr+1 = nCr xn-r . yr
- General Term in (1 + x)n is nCr xr
- In the binomial expansion of (x + y)n , the rth term from end is (n – r + 2)th .