Question

In the expansion of ${(x+\sqrt{x^2-1})}^6$+ ${(x-\sqrt{x^2-1})}^6$,the number of terms is

• A. $7$
• B. $14$
• C. $6$
• D. $4$

Answer 1

The correct option is D $4$

Let $a=x,b=\sqrt{x^2-1}$
Now,
$(x+\sqrt{x^2-1}{)}^6=(a+b{)}^6$
${=}^6{C}_0{a}^6{+}^6{C}_1{a}^{5}b{+}^6{C}_2{a}^4{b}^2{+}^6{C}_3{a}^3{b}^3{+}^6{C}_4{a}^2{b}^4{+}^6{C}_{5}a{b}^5{+}^6{C}_6{b}^6$
And $(x-\sqrt{x^2-1}{)}^6=(a-b{)}^6$
${=}^6{C}_0{a}^6{-}^6{C}_1{a}^{5}b{+}^6{C}_2{a}^4{b}^2{-}^6{C}_3{a}^3{b}^3{+}^6{C}_4{a}^2{b}^4{-}^6{C}_{5}a{b}^5{+}^6{C}_6{b}^6$
Now adding above two equations,
$(a+b{)}^6+(a-b{)}^6=2{(}^6{C}_0{a}^6{+}^6{C}_2{a}^4{b}^2{+}^6{C}_4{a}^2{b}^4{+}^6{C}_6{b}^6)$
Hence number of terms in the required expansion is $4$