Single Choice

If $$n$$ is even integer and $$a,b,c$$ are distinct, the number of distinct terms in the expansion of $${ \left( a+b+c \right) }^{ n }+{ \left( a+b-c \right) }^{ n }$$ is

A$$\displaystyle { \left( \frac { n }{ 2 } \right) }^{ 2 }$$
B$$\displaystyle { \left( \frac { n+1 }{ 2 } \right) }^{ 2 }$$
C$$\displaystyle { \left( \frac { n+2 }{ 2 } \right) }^{ 2 }$$
Correct Answer
D$$\displaystyle { \left( \frac { n+3 }{ 2 } \right) }^{ 2 }$$

Solution

Let $$n=2m,m\in N$$
$$\therefore { \left( a+b+c \right) }^{ n }+{ \left( a+b-c \right) }^{ n }={ \left[ \left( a+b \right) +c \right] }^{ 2m }+{ \left[ \left( a+b \right) -c \right] }^{ 2m }\\ =2\left\{ { \left( a+b \right) }^{ 2m }+^{ 2m }{ { C }_{ 2 }{ \left( a+b \right) }^{ 2m-2 }{ c }^{ 2 } }+...+^{ 2m }{ { C }_{ 2m }{ c }^{ 2m } } \right\} $$
Therefore, the number of distinct terms in the expansion
$$\displaystyle =\left( 2m+1 \right) +\left( 2m-1 \right) +...+3+1=\left( \frac { m+1 }{ 1 } \right) .\left( 2m+1+1 \right) $$
$$\displaystyle ={ \left( m+1 \right) }^{ 2 }={ \left( 1+\frac { n }{ 2 } \right) }^{ 2 }={ \left( \frac { n+2 }{ 2 } \right) }^{ 2 }$$

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