JEE Main 2026 — Mathematics Algebra

Let $\displaystyle S = \displaystyle\sum_{n=1}^{k} f(n)$S=n=1∑k​f(n), where

$$f(n) = \begin{vmatrix} n & -1 & -5 \\ -2n^2 & 3(2k+1) & 2k+1 \\ -3n^3 & 3k(2k+1) & 3k(k+2)+1 \end{vmatrix}$$f(n)=​n−2n2−3n3​−13(2k+1)3k(2k+1)​−52k+13k(k+2)+1​​

Since only the first column depends on $n$n, the summation can be taken inside that column:

$$S = \begin{vmatrix} \displaystyle\sum_{n=1}^{k} n & -1 & -5 \\ \displaystyle\sum_{n=1}^{k} (-2n^2) & 3(2k+1) & 2k+1 \\ \displaystyle\sum_{n=1}^{k} (-3n^3) & 3k(2k+1) & 3k(k+2)+1 \end{vmatrix}$$S=​n=1∑k​nn=1∑k​(−2n2)n=1∑k​(−3n3)​−13(2k+1)3k(2k+1)​−52k+13k(k+2)+1​​

Using standard summation formulas:

$$\displaystyle\sum_{n=1}^{k} n = \dfrac{k(k+1)}{2}$$n=1∑k​n=2k(k+1)​

$$\displaystyle\sum_{n=1}^{k} (-2n^2) = -\dfrac{k(k+1)(2k+1)}{3}$$n=1∑k​(−2n2)=−3k(k+1)(2k+1)​

$$\displaystyle\sum_{n=1}^{k} (-3n^3) = -\dfrac{3k^2(k+1)^2}{4}$$n=1∑k​(−3n3)=−43k2(k+1)2​

Substituting:

$$S = \begin{vmatrix} \dfrac{k(k+1)}{2} & -1 & -5 \\ -\dfrac{k(k+1)(2k+1)}{3} & 3(2k+1) & 2k+1 \\ -\dfrac{3k^2(k+1)^2}{4} & 3k(2k+1) & 3k^2+6k+1 \end{vmatrix}$$S=​2k(k+1)​−3k(k+1)(2k+1)​−43k2(k+1)2​​−13(2k+1)3k(2k+1)​−52k+13k2+6k+1​​

Taking $\dfrac{k(k+1)}{12}$12k(k+1)​ common from $C_1$C1​:

$$S = \dfrac{k(k+1)}{12} \begin{vmatrix} 6 & -1 & -5 \\ -4(2k+1) & 3(2k+1) & 2k+1 \\ -9k(k+1) & 3k(2k+1) & 3k^2+6k+1 \end{vmatrix}$$S=12k(k+1)​​6−4(2k+1)−9k(k+1)​−13(2k+1)3k(2k+1)​−52k+13k2+6k+1​​

Applying $C_1 \to C_1 + C_2 + C_3$C1​→C1​+C2​+C3​:

Row 1: $6 - 1 - 5 = 0$6−1−5=0

Row 2: $-8k - 4 + 6k + 3 + 2k + 1 = 0$−8k−4+6k+3+2k+1=0

Row 3: $-9k^2 - 9k + 6k^2 + 3k + 3k^2 + 6k + 1 = 1$−9k2−9k+6k2+3k+3k2+6k+1=1

$$S = \dfrac{k(k+1)}{12} \begin{vmatrix} 0 & -1 & -5 \\ 0 & 3(2k+1) & 2k+1 \\ 1 & 3k(2k+1) & 3k^2+6k+1 \end{vmatrix}$$S=12k(k+1)​​001​−13(2k+1)3k(2k+1)​−52k+13k2+6k+1​​

Expanding along $C_1$C1​:

$$S = \dfrac{k(k+1)}{12} \cdot 1 \cdot \begin{vmatrix} -1 & -5 \\ 3(2k+1) & 2k+1 \end{vmatrix}$$S=12k(k+1)​⋅1⋅​−13(2k+1)​−52k+1​​

$$S = \dfrac{k(k+1)}{12}\left[(-1)(2k+1) - (-5)(3(2k+1))\right]$$S=12k(k+1)​[(−1)(2k+1)−(−5)(3(2k+1))]

$$S = \dfrac{k(k+1)}{12}\left[-(2k+1) + 15(2k+1)\right]$$S=12k(k+1)​[−(2k+1)+15(2k+1)]

$$S = \dfrac{k(k+1)}{12} \cdot 14(2k+1) = 7 \cdot \dfrac{k(k+1)(2k+1)}{6}$$S=12k(k+1)​⋅14(2k+1)=7⋅6k(k+1)(2k+1)​

Note that $\displaystyle \dfrac{k(k+1)(2k+1)}{6} = \displaystyle\sum_{n=1}^{k} n^2$6k(k+1)(2k+1)​=n=1∑k​n2.

Given $S = 98$S=98:

$$7 \cdot \dfrac{k(k+1)(2k+1)}{6} = 98$$7⋅6k(k+1)(2k+1)​=98

$$\dfrac{k(k+1)(2k+1)}{6} = 14$$6k(k+1)(2k+1)​=14

Checking integer values of $k$k:

$k = 1$k=1: sum $= 1$=1

$k = 2$k=2: sum $= 5$=5

$k = 3$k=3: sum $= 14$=14

Hence, $k = 3$k=3, and the correct option is $(1)\ 3$(1) 3.