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Question: If \[1 + {x^4} + {x^5} = \sum\limits_{i = 0}^5 {{a_i}{{(1 + x)}^i}} \] for all x in R, then \[{a_4}\...

If 1+x4+x5=∑i=05ai(1+x)i1 + {x^4} + {x^5} = \sum\limits_{i = 0}^5 {{a_i}{{(1 + x)}^i}} for all x in R, then a4{a_4}is
A. −4 - 4
B. 66
C. −8 - 8
D. 1010

Explanation

Solution

Here we substitute the values of i in RHS of the equation from 1 to 5 and write the sum of terms. Then we start differentiating both sides of the equation which will give us one term constant, similarly we differentiate both the sides until we get the constant term on the right hand side as a4{a_4}.

Formula used: Differentiation is done by formula d(xn)dx=nxn−1\dfrac{{d({x^n})}}{{dx}} = n{x^{n - 1}}

Complete step-by-step answer:
We are given in the question 1+x4+x5=∑i=05ai(1+x)i1 + {x^4} + {x^5} = \sum\limits_{i = 0}^5 {{a_i}{{(1 + x)}^i}} … (1)
First we will open the terms on the right hand side of the equation.
We have RHS of the equation as ∑i=05ai(1+x)i\sum\limits_{i = 0}^5 {{a_i}{{(1 + x)}^i}} . We know that the summation sign opens up as sum of terms where the values of i=0,1,2,3,4,5i = 0,1,2,3,4,5
Therefore, ∑i=05ai(1+x)i=a0(1+x)0+a1(1+x)1+a2(1+x)2+a3(1+x)3+a4(1+x)4+a5(1+x)5\sum\limits_{i = 0}^5 {{a_i}{{(1 + x)}^i}} = {a_0}{(1 + x)^0} + {a_1}{(1 + x)^1} + {a_2}{(1 + x)^2} + {a_3}{(1 + x)^3} + {a_4}{(1 + x)^4} + {a_5}{(1 + x)^5}
Since, we know that any number having power as zero becomes one, i.e.p0=1{p^0} = 1
∴(1+x)0=1\therefore {(1 + x)^0} = 1
RHS of the equation becomes
a0+a1(1+x)+a2(1+x)2+a3(1+x)3+a4(1+x)4+a5(1+x)5{a_0} + {a_1}(1 + x) + {a_2}{(1 + x)^2} + {a_3}{(1 + x)^3} + {a_4}{(1 + x)^4} + {a_5}{(1 + x)^5}
Now we substitute RHS in equation (1)
1+x4+x5=a0+a1(1+x)+a2(1+x)2+a3(1+x)3+a4(1+x)4+a5(1+x)51 + {x^4} + {x^5} = {a_0} + {a_1}(1 + x) + {a_2}{(1 + x)^2} + {a_3}{(1 + x)^3} + {a_4}{(1 + x)^4} + {a_5}{(1 + x)^5}
Now we differentiate both sides with respect to x
⇒ddx[1+x4+x5]=ddx[a0+a1(1+x)+a2(1+x)2+a3(1+x)3+a4(1+x)4+a5(1+x)5]\Rightarrow \dfrac{d}{{dx}}[1 + {x^4} + {x^5}] = \dfrac{d}{{dx}}[{a_0} + {a_1}(1 + x) + {a_2}{(1 + x)^2} + {a_3}{(1 + x)^3} + {a_4}{(1 + x)^4} + {a_5}{(1 + x)^5}]
Substitute differentiation of constant terms as zero and use the formula for differentiation d(xn)dx=nxn−1\dfrac{{d({x^n})}}{{dx}} = n{x^{n - 1}} for all other terms
⇒4x3+5x4=a1+2a2(1+x)+3a3(1+x)2+4a4(1+x)3+5a5(1+x)4\Rightarrow 4{x^3} + 5{x^4} = {a_1} + 2{a_2}(1 + x) + 3{a_3}{(1 + x)^2} + 4{a_4}{(1 + x)^3} + 5{a_5}{(1 + x)^4}
Here the constant term on the right side is a1{a_1}, but we need a4{a_4}
Now differentiate again on both sides of the equation
⇒ddx[4x3+5x4]=ddx[a1+2a2(1+x)+3a3(1+x)2+4a4(1+x)3+5a5(1+x)4]\Rightarrow \dfrac{d}{{dx}}[4{x^3} + 5{x^4}] = \dfrac{d}{{dx}}[{a_1} + 2{a_2}(1 + x) + 3{a_3}{(1 + x)^2} + 4{a_4}{(1 + x)^3} + 5{a_5}{(1 + x)^4}]
Substitute differentiation of constant terms as zero and use the formula for differentiation d(xn)dx=nxn−1\dfrac{{d({x^n})}}{{dx}} = n{x^{n - 1}} for all other terms

⇒4×3x2+5×4x3=2a2+3a3×2(1+x)+4a4×3(1+x)2+5a5×4(1+x)3] ⇒12x2+20x3=2a2+6a3(1+x)+12a4(1+x)2+20a5(1+x)3  \Rightarrow 4 \times 3{x^2} + 5 \times 4{x^3} = 2{a_2} + 3{a_3} \times 2(1 + x) + 4{a_4} \times 3{(1 + x)^2} + 5{a_5} \times 4{(1 + x)^3}] \\\ \Rightarrow 12{x^2} + 20{x^3} = 2{a_2} + 6{a_3}(1 + x) + 12{a_4}{(1 + x)^2} + 20{a_5}{(1 + x)^3} \\\

Here the constant term on the right side is a2{a_2}, but we need a4{a_4}
Now differentiate again on both sides of the equation
⇒ddx[12x2+20x3]=ddx[2a2+6a3(1+x)+12a4(1+x)2+20a5(1+x)3]\Rightarrow \dfrac{d}{{dx}}[12{x^2} + 20{x^3}] = \dfrac{d}{{dx}}[2{a_2} + 6{a_3}(1 + x) + 12{a_4}{(1 + x)^2} + 20{a_5}{(1 + x)^3}]
Substitute differentiation of constant terms as zero and use the formula for differentiation d(xn)dx=nxn−1\dfrac{{d({x^n})}}{{dx}} = n{x^{n - 1}} for all other terms

⇒12×2x+20×3x2=6a3+12a4×2(1+x)+20a5×3(1+x)2 ⇒24x+60x2=6a3+24a4(1+x)+60a5(1+x)2  \Rightarrow 12 \times 2x + 20 \times 3{x^2} = 6{a_3} + 12{a_4} \times 2(1 + x) + 20{a_5} \times 3{(1 + x)^2} \\\ \Rightarrow 24x + 60{x^2} = 6{a_3} + 24{a_4}(1 + x) + 60{a_5}{(1 + x)^2} \\\

Here the constant term on the right side is a3{a_3}, but we need a4{a_4}
Now differentiate again on both sides of the equation
⇒ddx[24x+60x2]=ddx[6a3+24a4(1+x)+60a5(1+x)2]\Rightarrow \dfrac{d}{{dx}}[24x + 60{x^2}] = \dfrac{d}{{dx}}[6{a_3} + 24{a_4}(1 + x) + 60{a_5}{(1 + x)^2}]
Substitute differentiation of constant terms as zero and use the formula for differentiation d(xn)dx=nxn−1\dfrac{{d({x^n})}}{{dx}} = n{x^{n - 1}} for all other terms

⇒24+60×2x=24a4+60a5×2(1+x) ⇒24+120x=24a4+120a5(1+x)  \Rightarrow 24 + 60 \times 2x = 24{a_4} + 60{a_5} \times 2(1 + x) \\\ \Rightarrow 24 + 120x = 24{a_4} + 120{a_5}(1 + x) \\\

Now we have the constant term a4{a_4} on RHS of the equation, therefore, we solve the equation by substituting x=−1x = - 1 on both sides of the equation as it will make the bracket equal to zero

⇒24+120(−1)=24a4+120a5(1−1) ⇒24−120=24a4+0 ⇒−96=24a4  \Rightarrow 24 + 120( - 1) = 24{a_4} + 120{a_5}(1 - 1) \\\ \Rightarrow 24 - 120 = 24{a_4} + 0 \\\ \Rightarrow - 96 = 24{a_4} \\\

Divide both sides by 24
⇒−9624=24a424\Rightarrow \dfrac{{ - 96}}{{24}} = \dfrac{{24{a_4}}}{{24}}
Cancel out same factors from numerator and denominator
⇒−4=a4\Rightarrow - 4 = {a_4}

So, the correct answer is “Option A”.

Note: Students mostly make the mistake of substituting the value of x as -1 after the very first differentiation because they eventually try to bring a4{a_4} on one side of the equation but that will not give us a numeric value.