Given that x1x2x3x4xn=211,=375,=420,=523,and=xn−1−xn−2+xn−3−xn−4whenn≥5,find the value of x531+x753+x975.
Solution — click to reveal
Shifting the recurrence over by one and adding, we have: xnxn−1⟹xn+xn−1=xn−1−xn−2+xn−3−xn−4=xn−2−xn−3+xn−4−xn−5=xn−1−xn−5so xn=−xn−5 for all n. In particular, xn=−xn−5=−(−xn−10)=xn−10, so the sequence repeats with period 10. Thus, x531+x753+x975=x1+x3+x5=x1+x3+(x4−x3+x2−x1)=x2+x4=375+523=898.