Loading [MathJax]/jax/output/SVG/jax.js
 
+0  
 
+1
1062
2
avatar+1713 

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 113+135+157+179++1199201? Express your answer as a fraction in simplest form.

 

Please Help! Again, I am sorry I am posting 2 times in a row... 

 

😫😩😢😭😰😥😓

 Jul 10, 2019
 #1
avatar+6252 
+1

The formula seems to be nk=11(2k1)(2k+1)=n2n+1

 

Let's see if we can prove this by induction

 

P1:113=12(1)+1=TrueAssume Pn and prove PnPn+1 Let Sn=nk=11(2k1)(2k+1)Sn+1=Sn+1(2n+1)(2n+3)=n2n+1+1(2n+1)(2n+3)=12n+1(n+12n+3)=

 

12n+12n2+3n+12n+3=n+12n+3=n+12(n+1)+1and thus PnPn+1

 Jul 10, 2019
 #2
avatar+26397 
+3

What is the value of the sum113+135+157+179++1199201?

Express your answer as a fraction in simplest form.

 

113+135+157+179++1199201=113+135+157+179++1(2n1)(2n+1)1(2n1)(2n+1)=12(12n112n+1)113=12(1113)135=12(1315)157=12(1517)179=12(1719)1199201=12(11991201)=12(1113)+12(1315)+12(1517)+12(1719)++12(11991201)=12(1113+13=015+15=017+17=019+19=0+1199+1199=01201)=12(111201)=12(11201)=12(2011201)=12(200201)=100201

 

laugh

 Jul 10, 2019

1 Online Users

avatar