Question

The sum of coefficient of integral powers of x in the binomial expansion of ${{\left( 1-2\sqrt{x} \right)}^{50}}$ .
( a ) $\dfrac{{{3}^{50}}+1}{2}$
( b ) $\dfrac{{{3}^{50}}}{2}$
( c ) $\dfrac{{{3}^{50}}-1}{2}$
( d ) $\dfrac{{{2}^{50}}+1}{2}$

Answer 1

Hint: To solve this question what we will do is first we will expand the ${{\left( 1-2\sqrt{x} \right)}^{50}}$, then we will take the coefficients out, which will give the integral power of x and name that series as S, then we will find out the value of expansion of ${{\left( 1+x \right)}^{50}}$at x =2, -2 and then we will solve for value of S.

Complete step-by-step answer:
In this question, we are asked to find the sum of coefficients of integral power of x in the binomial expansion of ${{\left( 1-2\sqrt{x} \right)}^{50}}$ .
Now, firstly we see what is the binomial expansion of ${{(a+b)}^{n}}$ ,
So, ${{(a+b)}^{n}}={{a}^{n}}{{+}^{n}}{{C}_{1}}{{a}^{n-1}}b{{+}^{n}}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+.......{{+}^{n}}{{C}_{n-1}}a{{b}^{n-1}}+{{b}^{n}}$….( i )
Now, in ${{\left( 1-2\sqrt{x} \right)}^{50}}$, a = 1, $b=-2\sqrt{x}$ and n = 50.
So, substituting a = 1, $b=-2\sqrt{x}$ and n = 50 in ( i )
${{\left( 1-2\sqrt{x} \right)}^{50}}{{=}^{50}}{{C}_{0}}{{+}^{50}}{{C}_{1}}{{(-2\sqrt{x})}^{1}}{{+}^{50}}{{C}_{2}}{{(-2\sqrt{x})}^{2}}{{+}^{50}}{{C}_{3}}{{(-2\sqrt{x})}^{3}}{{+}^{50}}{{C}_{4}}{{(-2\sqrt{x})}^{4}}+.........$
On seeing expansion, we can see that terms with even powers will only have the integral power of x, as square root will vanish in even power and will give us only integral power.
So, Let sum of coefficients be denoted by S, then
$S{{=}^{50}}{{C}_{0}}{{+}^{50}}{{C}_{2}}{{(-2)}^{2}}{{+}^{50}}{{C}_{4}}{{(-2)}^{4}}{{+}^{50}}{{C}_{6}}{{(-2)}^{6}}+.........{{+}^{50}}{{C}_{50}}{{(-2)}^{50}}$
Now, if we expand ${{\left( 1+x \right)}^{50}}$ , using binomial expansion, we get
${{\left( 1+x \right)}^{50}}=1{{+}^{50}}{{C}_{1}}{{x}^{1}}{{+}^{50}}{{C}_{2}}{{(x)}^{2}}{{+}^{50}}{{C}_{3}}{{(x)}^{3}}{{+}^{50}}{{C}_{4}}{{(x)}^{4}}{{+}^{50}}{{C}_{5}}{{(x)}^{5}}{{+}^{50}}{{C}_{6}}{{(x)}^{6}}+.........{{+}^{50}}{{C}_{50}}{{(x)}^{50}}$.( ii )
Substituting x =2 in ( ii ) , we get
${{\left( 1+2 \right)}^{50}}=1{{+}^{50}}{{C}_{1}}{{2}^{1}}{{+}^{50}}{{C}_{2}}{{(2)}^{2}}{{+}^{50}}{{C}_{3}}{{(2)}^{3}}{{+}^{50}}{{C}_{4}}{{(2)}^{4}}{{+}^{50}}{{C}_{5}}{{(2)}^{5}}{{+}^{50}}{{C}_{6}}{{(2)}^{6}}+.........{{+}^{50}}{{C}_{50}}{{(2)}^{50}}$……( iii )
Substituting x = -2 in ( ii ), we get
${{\left( 1-2 \right)}^{50}}=1{{+}^{50}}{{C}_{1}}{{(-2)}^{1}}{{+}^{50}}{{C}_{2}}{{(-2)}^{2}}{{+}^{50}}{{C}_{3}}{{(-2)}^{3}}{{+}^{50}}{{C}_{4}}{{(-2)}^{4}}{{+}^{50}}{{C}_{5}}{{(-2)}^{5}}{{+}^{50}}{{C}_{6}}{{(-2)}^{6}}+.........{{+}^{50}}{{C}_{50}}{{(-2)}^{50}}$
${{\left( 1-2 \right)}^{50}}=1{{-}^{50}}{{C}_{1}}(2){{+}^{50}}{{C}_{2}}{{(2)}^{2}}{{-}^{50}}{{C}_{3}}{{(2)}^{3}}{{+}^{50}}{{C}_{4}}{{(2)}^{4}}{{-}^{50}}{{C}_{5}}{{(2)}^{5}}{{+}^{50}}{{C}_{6}}{{(2)}^{6}}+.........{{+}^{50}}{{C}_{50}}{{(2)}^{50}}$……( iv )
Adding ( iii ) and ( iv ), we get
${{3}^{50}}+1=2\left( ^{50}{{C}_{0}}{{+}^{50}}{{C}_{2}}{{(-2)}^{2}}{{+}^{50}}{{C}_{4}}{{(-2)}^{4}}{{+}^{50}}{{C}_{6}}{{(-2)}^{6}}+.........{{+}^{50}}{{C}_{50}}{{(-2)}^{50}} \right)$
${{3}^{50}}+1=2\left( ^{50}{{C}_{0}}{{+}^{50}}{{C}_{2}}{{(2)}^{2}}{{+}^{50}}{{C}_{4}}{{(2)}^{4}}{{+}^{50}}{{C}_{6}}{{(2)}^{6}}+.........{{+}^{50}}{{C}_{50}}{{(2)}^{50}} \right)$,
As, $S{{=}^{50}}{{C}_{0}}{{+}^{50}}{{C}_{2}}{{(-2)}^{2}}{{+}^{50}}{{C}_{4}}{{(-2)}^{4}}{{+}^{50}}{{C}_{6}}{{(-2)}^{6}}+.........{{+}^{50}}{{C}_{50}}{{(-2)}^{50}}$
So, ${{3}^{50}}+1=2\left( S \right)$
So, $S=\dfrac{{{3}^{50}}+1}{2}$

So, the correct answer is “Option (a)”.

Note: While solving binomial expansion questions, always remember the expansion of ${{(a+b)}^{n}}$ which is equals to ${{(a+b)}^{n}}={{a}^{n}}{{+}^{n}}{{C}_{1}}{{a}^{n-1}}b{{+}^{n}}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+.......{{+}^{n}}{{C}_{n-1}}a{{b}^{n-1}}+{{b}^{n}}$. Solve questions strictly according to the need of the question. Calculation mistakes must be avoided.