The coefficient of $x^7$ in the expression $\left(1+x\right)^{10}+x\left(1+x\right)^{9}+x^{2}\left(1+x\right)^{8}+...+x^{10}$ is :
- $420$
- $330$
- $210$
- $120$
The Correct Option is B
Solution and Explanation
Top JEE Main Mathematics Questions
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- The general solution of
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Top JEE Main binomial expansion formula Questions
- Number of integral terms in the expansion of \( \left( \sqrt{7} z + \frac{1}{6 \sqrt{z}} \right)^{824} \) is equal to ______.
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- For some \( n \neq 10 \), let the coefficients of the 5th, 6th, and 7th terms in the binomial expansion of \( (1 + x)^{n+4} \) be in A.P. Then the largest coefficient in the expansion of \( (1 + x)^{n+4} \) is:
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
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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 .