Question

Find the seventh term of the expansion of

Answer 1

Answer:

The correct answer option is C. $462 (3x)^5 (7y)^6$.

Step-by-step explanation:

We are given the following expression to be expanded and we are to find its seventh term:

$(3x+7y)^{11}$

The coefficient here is taken from Pascal's triangle (nCr on the calculator).

The expression expands in a way such that, the power of the first term decreases by one each time while it increases for the second term.

1st term: $1 (3x)^11$

2nd term: $11 (3x)^{10} (7y)^1$

3rd term: $55 (3x)^9 (7y)^2$ and so on.

Therefore, the 7th term will be $462 (3x)^5 (7y)^6$

Answer 2

Answer:

Correct choice is C

Step-by-step explanation:

The i-th term of the binomial expansion $\left(3x+7y\right)^{11}$ is

$T_i=C^{11}_{i-1}\cdot \left(3x\right)^{11+1-i}\cdot (7y)^{i-1}.$

If i=7, then

$T_7=C^{11}_{7-1}\cdot \left(3x\right)^{11+1-7}\cdot (7y)^{7-1}=C_6^{11}\cdot (3x)^5\cdot (7y)^6=462\cdot (3x)^5\cdot (7y)^6.$

Note that

$C_6^{11}=\dfrac{11!}{6!(11-6)!}=\dfrac{6!\cdot 7\cdot 8\cdot 9\cdot 10\cdot 11}{6!\cdot 1\cdot 2\cdot 3\cdot 4\cdot 5}=462.$