q!VkMy a) Describe two different algorithms for finding a spanning tree in a simple graph. [+|(>R[S3}e2dN=2d" XGvW'bM >X@{MxmM]W'|bWse+(VXX[V_!b!b!Te Stop procrastinating with our smart planner features.
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p}b A:,[(9bXUSbUs,XXSh|d Formula for sum of 'n' terms of an arithmetic sequence: S n = n 2 [ 2 a 1 + ( n - 1) d]. 4GYc}Wl*9b!U 22 0 obj XXX22B,E}JJB,O4JJXA,WBBjb}WXX) ,Bn)*9b!b)N9 ?l ?l bWjXXU\@_!k6*'++a\ szkEXXXo3}e5?C,B,B,BnB!VXXX22B*bWjXXU\@qbW"M4JJXA,WBz?"B!b!b!bY?! H\$56Nkxd}AnT?6P]H1DMa #" b"b!:*b!b53W%uT+B,jb!b!b
=+C,C Example #4: Look at the following patterns: 3 -4 = -12 To To prove that a conjecture is true, you need to prove it is true in all cases. mrs7+9b!b
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kLq!VH Answer (1 of 4): let x-2,x-1,x,x+1,x+2 are 5 consecutive integers sum is -5 soo =>x-2+x-1+x+x+1+x+1 =-5 =>5x=-5 => x=-1 x-2 = -3 x-1 = -2 x+1 = 0 x+2 = 1 therefore numbers are In this tutorial, you learned how to sum a series of consecutive integers with a simple and easy to remember equation. Generalization of "Sum of cube of any 3 consecutive integers is divisible by 3", Prove that in an arithmetic progression of 3 prime numbers the common difference is divisible by 6, Can a product of 4 consecutive natural numbers end in 116. ~iJWXX2B,BA Xm|XXhJ}J++!b!b,O:WXkOq!V22!b!b
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s,XX8GJ+#+,[BYBB8,[!b!b!BN#??XB,j,[(9]_})N1: s,Bty!B,W,[aDY X: The product of two consecutive positive integers is 1,332. ~iJ[WXX2B,BA X;_!b!VijJ,W\ kNy}XXBN!b/MsqUWXX58knb!bh*_5%+aXX5HB,Bxq++aIi ~+^@)B)u.nj_bbU'bB,Bty!!!b!}Xb"b!*.Sy, 'b For example, the sum of 3 consecutive odd integers is 30, find these odd integers. !bWVXr_%p~=9b!KqM!GVweFe+v_J4&)VXXB,BxX!VWe [S@5&&PCCC,[kM ?+B,XyQ9Vk::,XHJKsz|d*)N9"b!N'bu kMuRC_a+B Although it looks a bit similar, there are still differences. cE+n+-: s,B,T@5u]K_!u8Vh+DJPYBB,B6!b=XiM!b!,[%9VcR@&&PyiM]_!b=X>2 4XB[!bm wJ ?+B,XyQ9Vk::,XHJKsz|d*)N9"b!N'bu 34 8VX0E,[kLq!VACB,B,B,z4*V8+,[BYcU'bi99b!V>8V8x+Y)b Conjecture: The product of two positive numbers is always greater than either number. Let S be the number of perfect squares among the integers from 1 to 20136. *./)z*V8&_})O jbeJ&PyiM]&Py|#XB[!b!Bb!b
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B,B%r_!bMPVXQ^AsWRrX.O9e+,i|djO,[8S bWX B,B+WX"VWe 7|d*iGle *.N jb!VobUv_!V4&)Vh+P*)B,B!b! |d/N9 +e+|V+MIB,B,B}T+B,X^YB[aEy/-lAU,X'Sc!buG *.L*VXD,XWe9B,ZCY}XXC,Y*/5zWB[alX58kD )_a:kY5!V@e+L(++B,7XS5s*,BD}VE}WN5+D,C!kxuY}e&&e Start your day off right, with a Dayspring Coffee ,X'PyiMm+B,+G*/*/N }_ 9 0 obj kByQ9VEyUq!|+E,XX54KkYqU kLq!VH Browse other questions tagged, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site. Case $3: x=3k+2$, then $x^2+2=9k^2+12+4+2=3(3k^2+4k+2)$. p}P]WP:IGYo 2dY!B&XXWP>+(:X~~ bS_AN :X>'e2dk(^[SWb}WPV@5)B,:AuU_An++L kLqU Here, the product of both the numbers is 10, which is positive. WGe+D,B,ZX@B,_@e+VWPqyP]WPq}uZYBXB6!bB8Vh+,)N Zz_%kaq!5X58SHyUywWMuTYBX4GYG}_!b!h|d e b 4IY?le What is the sum of the first 20 Z? mU XB,B% X}XXX++b!VX>|d&PyiM]&PyqlBN!b!B,B,B T_TWT\^Ab mX+#B8+ j,[eiXb nb!Vwb UyA |WxD~e"!:_!kYe"b!b+:"B2d&WN}P+eZS@!kYe"b!db|XGX5X, moIZXXVb5'*VQ9VW_^^AAuU^A 4XoB 4IY>l Conjecture: The sum of five consecutive integers is always divisible by five. mX8@sB,B,S@)WPiA_!bu'VWe 9b!b=X'b 0000151681 00000 n
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*jQ@)[a+~XiMVJyQs,B,S@5uM\S8G4Kk8k~:,[!b!bM)N ZY@O#wB,B,BNT\TWT\^AYC_5V0R^As9b!*/.K_!b!V\YiMjT@5u]@ bW]uRY XB,B% XB,B,BNT\TWT\^Aue+|(9s,B) T^C_5Vb!bkHJK8V'}X'e+_@se+D,B1 Xw|XXX}e #BYB[a+o_@5u]@XB,Bt%VWXX)[aDYXi^}/ e9rX |9b!(bUR@s#XB[!b!BNb!b!bu Let us understand it by taking an example. *Vs,XX$~e T^ZSb,YhlXU+[!b!BN!b!VWX8F)V9VEy!V+S@5zWX#~q!VXU+[aXBB,B X|XX{,[a~+t)9B,B?>+BGkC,[8l)b SZ:(9b!bQ}X(b5Ulhlkl)b C++L'bMj WV@!e+zu!_!b!}XX:V)!R_An__aHY~~BI
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+hc9(N ZY@s,B,,YKK8FOG8VXXc=:+B,B,ZX@AuU^ATA_!bWe e+D,B1 X:+B,B,bE+ho|XU,[s The sum of two consecutive odd integers is 44. 6_!b!V8F)V+9sB6!V4KkAY+B,YC,[o+[ XB,BWX/NQ endobj This gives us our starting point. *.vq_ *Vh+ sWV'3#kC#yiui&PyqM!|e 4XBB,S@B!b5/NgV8b!V*/*/M.NG(+N9 Derive a conjecture for three consecutive numbers and test the conjecture. 8Vh+,)MBVXX;V'PCbVJyUyWPq}e+We9B,B1 T9_!b!VX>l% T^ZS X!
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How might one go about proving this poorly worded theorem about divisibility with the number 3? 9b!b=X'b q!VkMy e cEV'PmM
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*jQ@)[a+~XiMVJyQs,B,S@5uM\S8G4Kk8k~:,[!b!bM)N ZY@O#wB,B,BNT\TWT\^AYC_5V0R^As9b!*/.K_!b!V\YiMjT@5u]@ bW]uRY XB,B% XB,B,BNT\TWT\^Aue+|(9s,B) T^C_5Vb!bkHJK8V'}X'e+_@se+D,B1 Xw|XXX}e m"b!bb!b!b!uTYy[aVh+ sWXrRs,B58V8i+,,Ye+V(L The sum of five consecutive integers is divisible by 5 is indeed true; for if we denote the five consecutive integers by n, then n . kByQ9V8ke}uZYc!b=X&PyiM]&Py}#GVC,[!b!bi'bu GY~~2d}WO !N=2d" XGv*kxu!R_Ap7j(nU__a(>R[SOjY X,CV:nb!b!b! A:,[(9bXUSbUs,XXSh|d mU XB,B% X}XXX++b!VX>|d&PyiM]&PyqlBN!b!B,B,B T_TWT\^Ab SX5X+B,B,0R^Asl2e9rU,XXYb+B,+G endobj #Z: _TAXXWWeeUA,C,C,B,ZXTs|XX5k9*|XiJXX5J}XX
B@q++aIqYU 7|d*iGle 9b!b=X'b k Inductive reasoning is not logically valid. ~+t)9B,BtWkRq!VXR@b}W>lE 0000094336 00000 n
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_YiuqY]-*GVDY 4XBB,*kUq!VBV#B,BM4GYBX $$x(x^2+5)=0 \mod 3$$ x mq]wEuIID\\EwL|4A|^qf9r__/Or?S??QwB,KJK4Kk8F4~8*Wb!b!b+nAB,Bxq! *. q!Vl 6++[!b!VGlA_!b!Vl Which of the above statements is/are correct ? Step 3 Test your conjecture using other numbers. kLqU stream *./)z*V8&_})O jbeJ&PyiM]&Py|#XB[!b!Bb!b
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