If f: N -> Z is defined by f(n) = | n -1 -5 | | -2n^2 3(2k+1) 2k+1 | | -3n^3 3k(2k+1) 3k(k+2)+1 | , k in N, and sum from n=1 to k of f(n) = 98, then k is equal to :

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q3 – JEE-MAIN 2026

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Home Maths DeterminantsQuestion #160

f : N \to Z

is defined by

f (n) = | \begin{array}{ccc} n & - 1 & - 5 \\ - 2 n^{2} & 3 (2 k + 1) & 2 k + 1 \\ - 3 n^{3} & 3 k (2 k + 1) & 3 k (k + 2) + 1 \end{array} |, k \in N,

and

\sum_{n = 1}^{k} f (n) = 98

, then

k

is equal to :

Clue 1 of 3: Determinant Simplification

Simplify the given determinant $f (n)$ by applying column operations to introduce zeros, making expansion easier.

🧠 Full Solution Path

● Video Walkthrough

Step 1: Simplify the Determinant

f (n)

✦ Active

Apply the column operation $C_{2} \to C_{2} - 3 C_{3}$ to introduce a zero in the second column. Then, expand the determinant along the second row (which contains a zero) to express $f (n)$ as a polynomial in $n$ .

f (n) = | \begin{array}{ccc} n & - 1 & - 5 \\ - 2 n^{2} & 3 (2 k + 1) & 2 k + 1 \\ - 3 n^{3} & 3 k (2 k + 1) & 3 k (k + 2) + 1 \end{array} |

Applying

C_{2} \to C_{2} - 3 C_{3}

f (n) = | \begin{array}{ccc} n & 14 & - 5 \\ - 2 n^{2} & 0 & 2 k + 1 \\ - 3 n^{3} & - 3 (k^{2} + 5 k + 1) & 3 k (k + 2) + 1 \end{array} |

Expanding along

R_{2}

f (n) = - (- 2 n^{2}) | \begin{array}{cc} 14 & - 5 \\ - 3 (k^{2} + 5 k + 1) & 3 k (k + 2) + 1 \end{array} | - (2 k + 1) | \begin{array}{cc} n & 14 \\ - 3 n^{3} & - 3 (k^{2} + 5 k + 1) \end{array} |

f (n) = 2 n^{2} [14 (3 k^{2} + 6 k + 1) - 15 (k^{2} + 5 k + 1)] - (2 k + 1) [- 3 n (k^{2} + 5 k + 1) + 42 n^{3}]

f (n) = 2 n^{2} [27 k^{2} + 9 k - 1] + 3 n (2 k + 1) (k^{2} + 5 k + 1) - 42 n^{3} (2 k + 1)

Rearranging terms,

f (n) = - 42 (2 k + 1) n^{3} + 2 (27 k^{2} + 9 k - 1) n^{2} + 3 (2 k + 1) (k^{2} + 5 k + 1) n

Step 2: Evaluate the Summation

\sum_{n = 1}^{k} f (n)

○ Expand

Use the standard summation formulas for powers of $n$ : $\sum_{n = 1}^{k} n = \frac{k (k + 1)}{2}$ , $\sum_{n = 1}^{k} n^{2} = \frac{k (k + 1) (2 k + 1)}{6}$ , and $\sum_{n = 1}^{k} n^{3} = {(\frac{k (k + 1)}{2})}^{2}$ .

\sum_{n = 1}^{k} f (n) = - 42 (2 k + 1) \sum_{n = 1}^{k} n^{3} + 2 (27 k^{2} + 9 k - 1) \sum_{n = 1}^{k} n^{2} + 3 (2 k + 1) (k^{2} + 5 k + 1) \sum_{n = 1}^{k} n

\sum_{n = 1}^{k} f (n) = - 42 (2 k + 1) {(\frac{k (k + 1)}{2})}^{2} + 2 (27 k^{2} + 9 k - 1) \frac{k (k + 1) (2 k + 1)}{6} + 3 (2 k + 1) (k^{2} + 5 k + 1) \frac{k (k + 1)}{2}

Step 3: Substitute and Solve for

k

○ Expand

Substitute the given options for $k$ into the summation expression and check which one equals $98$ . Let's test $k = 3$ (Option 1):

For

k = 3

2 k + 1 = 2 (3) + 1 = 7

k^{2} + 5 k + 1 = 3^{2} + 5 (3) + 1 = 9 + 15 + 1 = 25

27 k^{2} + 9 k - 1 = 27 (3^{2}) + 9 (3) - 1 = 27 (9) + 27 - 1 = 243 + 27 - 1 = 269

\sum_{n = 1}^{3} n = 1 + 2 + 3 = 6

\sum_{n = 1}^{3} n^{2} = 1^{2} + 2^{2} + 3^{2} = 1 + 4 + 9 = 14

\sum_{n = 1}^{3} n^{3} = 1^{3} + 2^{3} + 3^{3} = 1 + 8 + 27 = 36

Now substitute these values into the summation formula:

\sum_{n = 1}^{3} f (n) = - 42 (7) (36) + 2 (269) (14) + 3 (7) (25) (6)

= - 294 (36) + 538 (14) + 525 (6)

= - 10584 + 7532 + 3150

= - 10584 + 10682 = 98

Since the sum is $98$ for $k = 3$ , the correct value of $k$ is $3$ .

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Maths Question 3 – JEE-MAIN 2026

Clue 1 of 3: Determinant Simplification

🧠 Full Solution Path