To find the value of k k k ∑ n = 1 k f ( n ) \sum_{n=1}^{k} f(n) ∑ n = 1 k f ( n )
Since the variable n n n f ( n ) f(n) f ( n )
∑ n = 1 k f ( n ) = ∣ ∑ n = 1 k n − 1 − 5 − 2 ∑ n = 1 k n 2 3 ( 2 k + 1 ) 2 k + 1 − 3 ∑ n = 1 k n 3 3 k ( 2 k + 1 ) 3 k ( k + 2 ) + 1 ∣ \sum_{n=1}^{k} f(n) = \begin{vmatrix} \sum_{n=1}^{k} n & -1 & -5 \\ -2 \sum_{n=1}^{k} n^2 & 3(2k+1) & 2k+1 \\ -3 \sum_{n=1}^{k} n^3 & 3k(2k+1) & 3k(k+2)+1 \end{vmatrix} ∑ n = 1 k f ( n ) = ∑ n = 1 k n − 2 ∑ n = 1 k n 2 − 3 ∑ n = 1 k n 3 − 1 3 ( 2 k + 1 ) 3 k ( 2 k + 1 ) − 5 2 k + 1 3 k ( k + 2 ) + 1
Using the standard summation formulas:
∑ n = 1 k n = k ( k + 1 ) 2 \sum_{n=1}^{k} n = \frac{k(k+1)}{2} ∑ n = 1 k n = 2 k ( k + 1 ) ∑ n = 1 k n 2 = k ( k + 1 ) ( 2 k + 1 ) 6 \sum_{n=1}^{k} n^2 = \frac{k(k+1)(2k+1)}{6} ∑ n = 1 k n 2 = 6 k ( k + 1 ) ( 2 k + 1 ) ∑ n = 1 k n 3 = k 2 ( k + 1 ) 2 4 \sum_{n=1}^{k} n^3 = \frac{k^2(k+1)^2}{4} ∑ n = 1 k n 3 = 4 k 2 ( k + 1 ) 2
Substituting these sums into the first column gives:
∑ n = 1 k f ( n ) = ∣ k ( k + 1 ) 2 − 1 − 5 − k ( k + 1 ) ( 2 k + 1 ) 3 3 ( 2 k + 1 ) 2 k + 1 − 3 k 2 ( k + 1 ) 2 4 3 k ( 2 k + 1 ) 3 k 2 + 6 k + 1 ∣ \sum_{n=1}^{k} f(n) = \begin{vmatrix} \frac{k(k+1)}{2} & -1 & -5 \\ -\frac{k(k+1)(2k+1)}{3} & 3(2k+1) & 2k+1 \\ -\frac{3k^2(k+1)^2}{4} & 3k(2k+1) & 3k^2+6k+1 \end{vmatrix} ∑ n = 1 k f ( n ) = 2 k ( k + 1 ) − 3 k ( k + 1 ) ( 2 k + 1 ) − 4 3 k 2 ( k + 1 ) 2 − 1 3 ( 2 k + 1 ) 3 k ( 2 k + 1 ) − 5 2 k + 1 3 k 2 + 6 k + 1
Now, we simplify the determinant by factoring out common terms:
Factor out ( 2 k + 1 ) (2k+1) ( 2 k + 1 ) R 2 R_2 R 2
Factor out k ( k + 1 ) k(k+1) k ( k + 1 ) C 1 C_1 C 1
This yields:
∑ n = 1 k f ( n ) = k ( k + 1 ) ( 2 k + 1 ) ∣ 1 2 − 1 − 5 − 1 3 3 1 − 3 k ( k + 1 ) 4 3 k ( 2 k + 1 ) 3 k 2 + 6 k + 1 ∣ \sum_{n=1}^{k} f(n) = k(k+1)(2k+1) \begin{vmatrix} \frac{1}{2} & -1 & -5 \\ -\frac{1}{3} & 3 & 1 \\ -\frac{3k(k+1)}{4} & 3k(2k+1) & 3k^2+6k+1 \end{vmatrix} ∑ n = 1 k f ( n ) = k ( k + 1 ) ( 2 k + 1 ) 2 1 − 3 1 − 4 3 k ( k + 1 ) − 1 3 3 k ( 2 k + 1 ) − 5 1 3 k 2 + 6 k + 1
To clear the fractions in the first column, multiply R 1 R_1 R 1 2 2 2 R 2 R_2 R 2 3 3 3 R 3 R_3 R 3 4 4 4 2 × 3 × 4 = 24 2 \times 3 \times 4 = 24 2 × 3 × 4 = 24
∑ n = 1 k f ( n ) = k ( k + 1 ) ( 2 k + 1 ) 24 ∣ 1 − 2 − 10 − 1 9 3 − 3 k 2 − 3 k 24 k 2 + 12 k 12 k 2 + 24 k + 4 ∣ \sum_{n=1}^{k} f(n) = \frac{k(k+1)(2k+1)}{24} \begin{vmatrix} 1 & -2 & -10 \\ -1 & 9 & 3 \\ -3k^2-3k & 24k^2+12k & 12k^2+24k+4 \end{vmatrix} ∑ n = 1 k f ( n ) = 24 k ( k + 1 ) ( 2 k + 1 ) 1 − 1 − 3 k 2 − 3 k − 2 9 24 k 2 + 12 k − 10 3 12 k 2 + 24 k + 4
Now, let us evaluate the determinant:
Perform row operation R 2 → R 2 + R 1 R_2 \to R_2 + R_1 R 2 → R 2 + R 1 ∣ 1 − 2 − 10 0 7 − 7 − 3 k 2 − 3 k 24 k 2 + 12 k 12 k 2 + 24 k + 4 ∣ \begin{vmatrix} 1 & -2 & -10 \\ 0 & 7 & -7 \\ -3k^2-3k & 24k^2+12k & 12k^2+24k+4 \end{vmatrix} 1 0 − 3 k 2 − 3 k − 2 7 24 k 2 + 12 k − 10 − 7 12 k 2 + 24 k + 4
Factor out 7 7 7 R 2 R_2 R 2 = 7 ∣ 1 − 2 − 10 0 1 − 1 − 3 k 2 − 3 k 24 k 2 + 12 k 12 k 2 + 24 k + 4 ∣ = 7 \begin{vmatrix} 1 & -2 & -10 \\ 0 & 1 & -1 \\ -3k^2-3k & 24k^2+12k & 12k^2+24k+4 \end{vmatrix} = 7 1 0 − 3 k 2 − 3 k − 2 1 24 k 2 + 12 k − 10 − 1 12 k 2 + 24 k + 4
Perform column operation C 3 → C 3 + C 2 C_3 \to C_3 + C_2 C 3 → C 3 + C 2 = 7 ∣ 1 − 2 − 12 0 1 0 − 3 k 2 − 3 k 24 k 2 + 12 k 36 k 2 + 36 k + 4 ∣ = 7 \begin{vmatrix} 1 & -2 & -12 \\ 0 & 1 & 0 \\ -3k^2-3k & 24k^2+12k & 36k^2+36k+4 \end{vmatrix} = 7 1 0 − 3 k 2 − 3 k − 2 1 24 k 2 + 12 k − 12 0 36 k 2 + 36 k + 4
Expanding along the second row:
= 7 ⋅ 1 ⋅ ∣ 1 − 12 − 3 k 2 − 3 k 36 k 2 + 36 k + 4 ∣ = 7 \cdot 1 \cdot \begin{vmatrix} 1 & -12 \\ -3k^2-3k & 36k^2+36k+4 \end{vmatrix} = 7 ⋅ 1 ⋅ 1 − 3 k 2 − 3 k − 12 36 k 2 + 36 k + 4 = 7 [ ( 36 k 2 + 36 k + 4 ) − ( − 12 ) ( − 3 k 2 − 3 k ) ] = 7 \left[ (36k^2+36k+4) - (-12)(-3k^2-3k) \right] = 7 [ ( 36 k 2 + 36 k + 4 ) − ( − 12 ) ( − 3 k 2 − 3 k ) ] = 7 [ 36 k 2 + 36 k + 4 − ( 36 k 2 + 36 k ) ] = 7 × 4 = 28 = 7 \left[ 36k^2+36k+4 - (36k^2+36k) \right] = 7 \times 4 = 28 = 7 [ 36 k 2 + 36 k + 4 − ( 36 k 2 + 36 k ) ] = 7 × 4 = 28
Substitute this determinant value back into our expression for ∑ n = 1 k f ( n ) \sum_{n=1}^{k} f(n) ∑ n = 1 k f ( n ) ∑ n = 1 k f ( n ) = k ( k + 1 ) ( 2 k + 1 ) 24 × 28 = 7 6 k ( k + 1 ) ( 2 k + 1 ) \sum_{n=1}^{k} f(n) = \frac{k(k+1)(2k+1)}{24} \times 28 = \frac{7}{6} k(k+1)(2k+1) ∑ n = 1 k f ( n ) = 24 k ( k + 1 ) ( 2 k + 1 ) × 28 = 6 7 k ( k + 1 ) ( 2 k + 1 )
We are given that ∑ n = 1 k f ( n ) = 98 \sum_{n=1}^{k} f(n) = 98 ∑ n = 1 k f ( n ) = 98 7 6 k ( k + 1 ) ( 2 k + 1 ) = 98 \frac{7}{6} k(k+1)(2k+1) = 98 6 7 k ( k + 1 ) ( 2 k + 1 ) = 98 k ( k + 1 ) ( 2 k + 1 ) = 98 × 6 7 = 84 k(k+1)(2k+1) = 98 \times \frac{6}{7} = 84 k ( k + 1 ) ( 2 k + 1 ) = 98 × 7 6 = 84
For k = 3 k = 3 k = 3 3 × ( 3 + 1 ) × ( 2 ( 3 ) + 1 ) = 3 × 4 × 7 = 84 3 \times (3+1) \times (2(3)+1) = 3 \times 4 \times 7 = 84 3 × ( 3 + 1 ) × ( 2 ( 3 ) + 1 ) = 3 × 4 × 7 = 84
Thus, k = 3 k = 3 k = 3