Bobosharif S. answered • 03/08/18
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Mathematics/Statistics Tutor
Tn=n2-4n
Tn+1=(n+1)2-4(n+1)
Tn+1-Tn=(n+1)2-4(n+1)-(n2-4n)=(n+1)2-n2-4(n+1-n)
=(n+1-n)(n+1+n)-4=2n-3
Bobosharif S.
First Tn=n2-4n. Then we can write Tn+1=(n+1)2-4(n+1) . Just replace n in the expression for Tn to n+1. Next, we take difference Tn+1-Tn=|just plug in expressions for Tn+1 and Tn |=(n+1)2-4(n+1)-(n2-4n)=|the I groupped squared and linear parts|= (n+1)2-n2 -4(n+1)+4n=|Here I used a2-b2=(a-b)(a+b), for a=n+1 and b=n|=(n+1-n)(n+1+n)-4(n+1-n)= I hope the rest is clear.
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03/08/18
Zac H.
03/08/18