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MCQMediumJEE 2026Sum of Series

JEE Mathematics 2026 Question with Solution

The positive integer nn, for which the solutions of the equation x(x+2)+(x+2)(x+4)+⋯+(x+2n−2)(x+2n)=8n3x(x+2) + (x+2)(x+4) + \dots + (x+2n-2)(x+2n) = \frac{8n}{3} are two consecutive even integers, is :

  • A

    99

  • B

    33

  • C

    1212

  • D

    66

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given:

∑k=0n−1(x+2k)(x+2k+2)=8n3\sum_{k=0}^{n-1}(x+2k)(x+2k+2)=\frac{8n}{3}

Find: The positive integer nn for which the two roots are consecutive even integers.

From the given sum,

∑k=0n−1(x+2k)(x+2k+2)\sum_{k=0}^{n-1}(x+2k)(x+2k+2)

the general term is expanded as

(x+2k)(x+2k+2)=x2+(4k+2)x+4k(k+1)(x+2k)(x+2k+2)=x^2+(4k+2)x+4k(k+1)

Now summing from k=0k=0 to n−1n-1,

nx2+[4(n−1)n2+2n]x+4[(n−1)n(2n−1)6+(n−1)n2]=8n3nx^2+\left[4\frac{(n-1)n}{2}+2n\right]x+4\left[\frac{(n-1)n(2n-1)}{6}+\frac{(n-1)n}{2}\right]=\frac{8n}{3}

Dividing by nn,

x2+2nx+23(n−1)(2n−1)+2(n−1)−83=0x^2+2nx+\frac{2}{3}(n-1)(2n-1)+2(n-1)-\frac{8}{3}=0

Simplifying the constant term,

x2+2nx+4n2−123=0x^2+2nx+\frac{4n^2-12}{3}=0

Let the roots be α\alpha and β\beta. Since they are consecutive even integers,

∣α−β∣=2|\alpha-\beta|=2

Using

(α−β)2=(α+β)2−4αβ(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta

for the quadratic,

α+β=−2n,αβ=4n2−123\alpha+\beta=-2n, \qquad \alpha\beta=\frac{4n^2-12}{3}

Hence,

22=(−2n)2−4(4n2−123)2^2=(-2n)^2-4\left(\frac{4n^2-12}{3}\right)

So,

4=4n2−16n2−4834=4n^2-\frac{16n^2-48}{3} 12=12n2−16n2+4812=12n^2-16n^2+48 4n2=364n^2=36 n=3n=3

Therefore, the positive integer is 33, so the correct option is B.

Using root difference condition

Given: The resulting equation in xx is quadratic, and its roots are consecutive even integers.

Find: nn.

For a quadratic equation ax2+bx+c=0ax^2+bx+c=0, if the roots differ by 22, then the discriminant satisfies

D=22a2=4a2D=2^2a^2=4a^2

Here, after simplification the equation is

x2+2nx+4n2−123=0x^2+2nx+\frac{4n^2-12}{3}=0

so a=1a=1.

Therefore,

D=(2n)2−4⋅1⋅4n2−123=4D=(2n)^2-4\cdot 1\cdot \frac{4n^2-12}{3}=4 4n2−16n2−483=44n^2-\frac{16n^2-48}{3}=4

which gives

n=3n=3

Thus, the correct option is B.

Common mistakes

  • Students may treat consecutive even integers as roots differing by 11. This is incorrect because consecutive even integers always differ by 22. Use ∣α−β∣=2|\alpha-\beta|=2.

  • While summing the expression, students may expand (x+2k)(x+2k+2)(x+2k)(x+2k+2) incorrectly or miss the term 4k(k+1)4k(k+1). Expand carefully before applying summation formulas.

  • Some students use Vieta's formulas with wrong signs. For x2+2nx+4n2−123=0x^2+2nx+\frac{4n^2-12}{3}=0, the sum of roots is −2n-2n, not 2n2n.

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