If is defined by and , then k is equal to : [2026]

Question

If $f : N \to Z$ is defined by $f (n) = | \begin{matrix} n & - 1 & - 5 \\ - 2 n^{2} & 3 (2 k + 1) & 2 k + 1 \\ - 3 n^{3} & 3 k (2 k + 1) & 3 k (k + 2) + 1 \end{matrix} |, k \in N$ and $\sum_{n = 1}^{k} f (n) = 98$ , then k is equal to : [2026]

Smart Achievers · Accepted Answer

(1)

We have,

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

Now, $\sum_{} n = \frac{k (k + 1)}{2}, \sum_{} n^{2} = \frac{k (k + 1) (2 k + 1)}{6}$

and $\sum_{n = 1}^{k} f (n) = | \begin{matrix} \frac{k (k + 1)}{2} & - 1 & - 5 \\ - 2 \cdot \frac{k \cdot (k + 1) (2 k + 1)}{6} & 3 (2 k + 1) & 2 k + 1 \\ - 3 \cdot \frac{k^{2} {(k + 1)}^{2}}{4} & 3 k (2 k + 1) & 3 k (k + 2) + 1 \end{matrix} |$

$\Rightarrow \frac{7 k (k + 1) (2 l + 1)}{6} = 98$

$\Rightarrow k (k + 1) (2 k + 1) = 84$

$∴ k = 3$

Solution

Q.
If $f : N \to Z$ is defined by $f (n) = | \begin{matrix} n & - 1 & - 5 \\ - 2 n^{2} & 3 (2 k + 1) & 2 k + 1 \\ - 3 n^{3} & 3 k (2 k + 1) & 3 k (k + 2) + 1 \end{matrix} |, k \in N$ and $\sum_{n = 1}^{k} f (n) = 98$ , then k is equal to : [2026]

1 3

2 4

3 5

4 6

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Solution

Q. If f:N→Z is defined by f(n)=|n–1–5–2n23(2k+1)2k+1–3n33k(2k+1)3k(k+2)+1|, k∈N and ∑n=1kf(n)=98, then k is equal to : [2026]

1 3

2 4

3 5

4 6

Ans. (1) We have, ∑f(n)=|∑n–1–5–2∑n23(2k+1)2k+1–3∑n33k(2k+1)3k(k+2)+1|, k∈N Now, ∑n=k(k+1)2, ∑n2=k(k+1)(2k+1)6 and ∑n=1kf(n)=|k(k+1)2–1–5–2·k·(k+1)(2k+1)63(2k+1)2k+1–3·k2(k+1)243k(2k+1)3k(k+2)+1| ⇒ 7k(k+1)(2l+1)6=98 ⇒ k(k+1)(2k+1)=84 ∴ k=3

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Q.
If $f : N \to Z$ is defined by $f (n) = | \begin{matrix} n & - 1 & - 5 \\ - 2 n^{2} & 3 (2 k + 1) & 2 k + 1 \\ - 3 n^{3} & 3 k (2 k + 1) & 3 k (k + 2) + 1 \end{matrix} |, k \in N$ and $\sum_{n = 1}^{k} f (n) = 98$ , then k is equal to : [2026]