Ok, here's the drill, please give me a nudge, this is a HOMEWORK PROBLEM! Yayyyy! I need help! Boooooo!
What is the value of the sum 11⋅3+13⋅5+15⋅7+17⋅9+⋯+1199⋅201? Express your answer as a fraction in simplest form.
Please Help! Again, I am sorry I am posting 2 times in a row...
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The formula seems to be n∑k=11(2k−1)(2k+1)=n2n+1
Let's see if we can prove this by induction
P1:11⋅3=12(1)+1=TrueAssume Pn and prove Pn⇒Pn+1 Let Sn=n∑k=11(2k−1)(2k+1)Sn+1=Sn+1(2n+1)(2n+3)=n2n+1+1(2n+1)(2n+3)=12n+1(n+12n+3)=
12n+1⋅2n2+3n+12n+3=n+12n+3=n+12(n+1)+1and thus Pn⇒Pn+1
What is the value of the sum11⋅3+13⋅5+15⋅7+17⋅9+⋯+1199⋅201?
Express your answer as a fraction in simplest form.
11∗3+13∗5+15∗7+17∗9+…+1199∗201=11∗3+13∗5+15∗7+17∗9+…+1(2n−1)(2n+1)1(2n−1)(2n+1)=12(12n−1−12n+1)11∗3=12(11−13)13∗5=12(13−15)15∗7=12(15−17)17∗9=12(17−19)…1199∗201=12(1199−1201)=12(11−13)+12(13−15)+12(15−17)+12(17−19)+…+12(1199−1201)=12(11−13+13⏟=0−15+15⏟=0−17+17⏟=0−19+19⏟=0+…−1199+1199⏟=0−1201)=12(11−1201)=12(1−1201)=12(201−1201)=12(200201)=100201