If xk+1 = xk + 1/2 for k = 1, 2, …, n-1 and x1 = 1, find x1 + x2 + ⋯ + xn.
Explanation:
xk+1 = xk + 1/2 ∴ x1, x2, x3, ..., xn form an arithmetic progression with common difference d = 1/2 ∵ x1 = 1, first term = a = 1
Sum of n terms of an arithmetic progression = n/2 × [2a + (n-1)d]
= n/2 × [2(1) + (n-1) 1/2]
= n/2 × [2 + n/2 - 1/2]
= n/2 × [(n + 3)/2]
= (n2 + 3n)/4
Hence, option (d).
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