Answer: Option A) [tex]729c^6 + 1,458c^5d^2 + 1,215c^4d^4 + 540c^3d^6 + 135c^2d^8 + 18cd^{10} + d^{12}[/tex] is the correct expansion.
Explanation:
on applying binomial theorem, [tex](a+b)^n=\sum_{r=0}^{n} \frac{n!}{r!(n-r)!} a^{n-r} b^r[/tex]
Here a=3c, [tex]b=d^2[/tex] and n=6,
Thus, [tex](3c+d^2)^6=\sum_{r=0}^{6} \frac{6!}{r!(6-r)!} (3c)^{n-r} (d^2)^r[/tex]
⇒ [tex](3c+d^2)^6= \frac{6!}{(6-0)!0!} (3c)^{6-0}.(d^2)^0+\frac{6!}{(6-1)!1!} (3c)^{6-1}.(d^2)^1+\frac{6!}{(6-2)!2!} (3c)^{6-2}.(d^2)^2+\frac{6!}{(6-3)!3!} (3c)^{6-3}.(d^2)^3+\frac{6!}{(6-4)!4!} (3c)^{6-4}.(d^2)^4+\frac{6!}{(6-5)!5!} (3c)^{6-5}.(d^2)^5+\frac{6!}{(6-6)!6!} (3c)^{6-6}.(d^2)^6[/tex]
⇒[tex](3c+d^2)^6= \frac{6!}{(6-)!0!} (3c)^6.d^0+\frac{6!}{(5)!1!} (3c)^5.d^2+\frac{6!}{(4)!2!} (3c)^4.d^4+\frac{6!}{(6-3)!3!} (3c)^3.d^6+\frac{6!}{(2)!4!} (3c)^2.d^8+\frac{6!}{(1)!5!} (3c).d^{10}+\frac{6!}{(0)!6!} (3c)^0.d^{12}[/tex]
⇒[tex](3c+d^2)^6=(3c)^6.d^0+\frac{720}{120} (3c)^5.d^2+\frac{720}{48} (3c)^4.d^4+\frac{720}{36} (3c)^3.d^6+\frac{720}{48} (3c)^2.d^8+\frac{720}{120} (3c).d^{10}+.d^{12}[/tex]
⇒[tex](3c+d^2)^6=729c^6 + 1,458c^5d^2 + 1,215c^4d^4 + 540c^3d^6 + 135c^2d^8 + 18cd^{10} + d^{12}[/tex]
Answer:
Option (a) is correct.
[tex](3c+d^2)^6=729c^6+1458c^5d^2+1215c^4d^4+540c^3d^6+135c^2d^8+18cd^{10}+d^{12}[/tex]
Step-by-step explanation:
Given : [tex]\left(3c+d^2\right)^6[/tex]
We have to expand the given expression and choose the correct from the given options.
Consider the given expression [tex]\left(3c+d^2\right)^6[/tex]
Using binomial theorem ,
[tex]\left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i[/tex]
We have [tex]a=3c,\:\:b=d^2[/tex]
[tex]=\sum _{i=0}^6\binom{6}{i}\left(3c\right)^{\left(6-i\right)}\left(d^2\right)^i[/tex]
also, [tex]\binom{n}{i}=\frac{n!}{i!\left(n-i\right)!}[/tex]
for i = 0 , we have,
[tex]\frac{6!}{0!\left(6-0\right)!}\left(3c\right)^6d^2^0=729c^6[/tex]
for i = 1 , we have,
[tex]\frac{6!}{1!\left(6-1\right)!}\left(3c\right)^5d^2^1=1458c^5d^2[/tex]
for i = 2 , we have,
[tex]\frac{6!}{2!\left(6-2\right)!}\left(3c\right)^4d^2^2=1215c^4d^4[/tex]
for i = 3 , we have,
[tex]\frac{6!}{3!\left(6-3\right)!}\left(3c\right)^3d^2^3=540c^3d^6[/tex]
for i = 4 , we have,
[tex]\frac{6!}{4!\left(6-4\right)!}\left(3c\right)^2d^2^4=135c^2d^8[/tex]
for i = 5 , we have,
[tex]\frac{6!}{4!\left(6-4\right)!}\left(3c\right)^2d^2^4=18cd^{10}[/tex]
for i = 6 , we have,
[tex]\frac{6!}{6!\left(6-6\right)!}\left(3c\right)^0d^2^6=d^{12}[/tex]
Thus, adding all term together, we have,
[tex](3c+d^2)^6=729c^6+1458c^5d^2+1215c^4d^4+540c^3d^6+135c^2d^8+18cd^{10}+d^{12}[/tex]
Thus, Option (a) is correct.
