<p>The value of \(\displaystyle\sum_{i=1}^{n} \sum_{j=1}^{i} \sum_{k=1}^{j} 1 = 220\), then the value of \(n\) equals</p>
Step-by-Step Solution
Key Concept: Each triple nested sum counts lattice points (i,j,k) where 1≤k≤j≤i≤n. This equals the binomial coefficient C(n+2,3), which counts combinations of choosing 3 items from n+2 items.
<p><strong>Step 1:</strong> Evaluate the innermost sum:<br>∑<sub>k=1</sub><sup>j</sup> 1 = j</p><p><strong>Step 2:</strong> Evaluate the middle sum:<br>∑<sub>j=1</sub><sup>i</sup> j = i(i+1)/2</p><p><strong>Step 3:</strong> Evaluate the outer sum:<br>∑<sub>i=1</sub><sup>n</sup> i(i+1)/2 = (1/2)∑<sub>i=1</sub><sup>n</sup> (i² + i)<br>= (1/2)[n(n+1)(2n+1)/6 + n(n+1)/2]<br>= (1/2)·n(n+1)[(2n+1)/6 + 1/2]<br>= (1/2)·n(n+1)[(2n+1+3)/6]<br>= n(n+1)(n+2)/6 = C(n+2,3)</p><p><strong>Step 4:</strong> Solve C(n+2,3) = 220<br>n(n+1)(n+2)/6 = 220<br>n(n+1)(n+2) = 1320</p><p><strong>Step 5:</strong> Test values:<br>For n = 10: 10·11·12 = 1320 ✓</p><p>∴ Answer: n = 10 (Option C)</p>
Correct Answer: C