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Question Number 6599 by WAI LIN last updated on 05/Jul/16

P r o v e t h a t \underset{j = 1}{\sum^{N}} {(X_{j} - 1)}^{2} = \underset{j = 1}{\sum^{N}} X_{j}^{2} - 2 \underset{j = 1}{\sum^{N}} X_{j} + N .

Answered by Temp last updated on 05/Jul/16

S = \underset{j = 1}{\sum^{n}} {(X_{j} - 1)}^{2}

S = \underset{j = 1}{\sum^{n}} (X_{j}^{2} - 2 X_{j} + 1)

S = \underset{j = 1}{\sum^{n}} X_{j}^{2} - \underset{j = 1}{\sum^{n}} 2 X_{j} + \underset{j = 1}{\sum^{n}} 1

S = \underset{j = 1}{\sum^{n}} X_{j}^{2} - 2 \underset{j = 1}{\sum^{n}} X_{j} + n

Q . E . D .

Commented by WAI LIN last updated on 05/Jul/16

T h a n k!

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