Which of the following is the expansion of $\left(3 c+d^2\right)^6$ ?

A$729 c^6+243 c^5 d^2+81 c^4 d^4+27 c^3 d^6+9 c^2 d^8+3 c d^{10}+d^{12}$
B$729 c^6+1,458 c^5 d+1,215 c^4 d^2+540 c^3 d^3+135 c^2 d^4+18 c d^5+d^6$
C$c^6+6 c^5 d^2+15 c^4 d^4+20 c^3 d^6+15 c^2 d^8+6 c d^{10}+d^{12}$
D$729 c^6+1,215 c^5 d^2+810 c^4 d^4+270 c^3 d^6+90 c^2 d^8+15 c d^{10}+d^{12}$
E$729 c^6+1,458 c^5 d^2+1,215 c^4 d^4+540 c^3 d^6+135 c^2 d^8+18 c d^{10}+d^{12}$

Question

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05-11-2026
Mathematics

Answered

Which of the following is the expansion of $\left(3 c+d^2\right)^6$ ?

A$729 c^6+243 c^5 d^2+81 c^4 d^4+27 c^3 d^6+9 c^2 d^8+3 c d^{10}+d^{12}$
B$729 c^6+1,458 c^5 d+1,215 c^4 d^2+540 c^3 d^3+135 c^2 d^4+18 c d^5+d^6$
C$c^6+6 c^5 d^2+15 c^4 d^4+20 c^3 d^6+15 c^2 d^8+6 c d^{10}+d^{12}$
D$729 c^6+1,215 c^5 d^2+810 c^4 d^4+270 c^3 d^6+90 c^2 d^8+15 c d^{10}+d^{12}$
E$729 c^6+1,458 c^5 d^2+1,215 c^4 d^4+540 c^3 d^6+135 c^2 d^8+18 c d^{10}+d^{12}$

Verified Answer :

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GinnyAnswer GinnyAnswer · Answer 1 · 2026-11-05T14:26:22-05:00

Apply the binomial theorem to $(3c + d^2)^6$ with $A=3c$, $B=d^2$, and $n=6$.
Calculate the seven terms using binomial coefficients $\binom{6}{k}$ for $k=0$ through $6$.
Sum the terms to obtain the full expansion: $729 c^6 + 1458 c^5 d^2 + 1215 c^4 d^4 + 540 c^3 d^6 + 135 c^2 d^8 + 18 c d^{10} + d^{12}$.
The correct option is $\boxed{e}$.

Explanation

Applying the Binomial Theorem

To expand the expression $(3c + d^2)^6$, we apply the Binomial Theorem, which states that for any non-negative integer $n$:
\[(A + B)^n = \sum_{k=0}^{n} \binom{n}{k} A^{n-k} B^k\]
In this problem, $A = 3c$, $B = d^2$, and $n = 6$. We will calculate each term for $k = 0, 1, 2, 3, 4, 5, 6$. The binomial coefficients $\binom{6}{k}$ are the numbers in the seventh row of Pascal's triangle: $1, 6, 15, 20, 15, 6, 1$.

Calculating Each Term

We calculate each of the seven terms individually:

$k=0$: $\binom{6}{0} (3c)^6 (d^2)^0 = 1 \times 729 c^6 \times 1 = 729 c^6$
$k=1$: $\binom{6}{1} (3c)^5 (d^2)^1 = 6 \times 243 c^5 \times d^2 = 1458 c^5 d^2$
$k=2$: $\binom{6}{2} (3c)^4 (d^2)^2 = 15 \times 81 c^4 \times d^4 = 1215 c^4 d^4$
$k=3$: $\binom{6}{3} (3c)^3 (d^2)^3 = 20 \times 27 c^3 \times d^6 = 540 c^3 d^6$
$k=4$: $\binom{6}{4} (3c)^2 (d^2)^4 = 15 \times 9 c^2 \times d^8 = 135 c^2 d^8$
$k=5$: $\binom{6}{5} (3c)^1 (d^2)^5 = 6 \times 3 c \times d^{10} = 18 c d^{10}$
$k=6$: $\binom{6}{6} (3c)^0 (d^2)^6 = 1 \times 1 \times d^{12} = d^{12}$

Finalizing the Expansion and Selecting the Correct Option

Combining all the calculated terms, we get the complete expansion:
\[729 c^6 + 1458 c^5 d^2 + 1215 c^4 d^4 + 540 c^3 d^6 + 135 c^2 d^8 + 18 c d^{10} + d^{12}\]
Comparing this result to the given options, it matches option (e).

Examples

The Binomial Theorem is a powerful tool in various fields, such as finance and probability. For instance, if you are calculating compound interest over multiple periods or determining the probability of a specific outcome in a series of independent events (like a sequence of coin flips), you often encounter binomial expressions. Expanding these expressions allows for a clearer understanding of the individual contributions of each variable or parameter, facilitating more accurate modeling and forecasting in real-world systems.