Tilted Forum Project Discussion Community  

Go Back   Tilted Forum Project Discussion Community > The Academy > Tilted Knowledge and How-To


 
 
LinkBack Thread Tools
Old 09-09-2003, 05:51 PM   #1 (permalink)
Upright
 
Discrete Math Help

Hello everyone, I was wondering if anyone here could help me out with a problem I'm trying to solve involving discrete math. The problem is an arithmatic-geometric series and I need to find the formula for the summation of it. Here it is:

The sum (sigma) from i=1 to n of [ i^2 * 3^i ]

If someone could help me, or show me how to solve this, I'd greatly appreciate it, Thanks!
sieger35 is offline  
Old 09-10-2003, 10:30 PM   #2 (permalink)
Insane
 
Location: Bay Area
Ah I just had Discrete Math last Winter and I don't even remember how to do this stuff. I sold my book and I do not have my notes.

I think you can start about by breaking it into two pieces:

for i = 1 to n sigma(i^2) * sigma(3^i)

Then there should be some general rule for what to do with sigma(i^x) and sigma(x^i)... giving you a couple formulas to multiply. And that is what I would look up in my text book if I had it.

Damn this is probably why I barely got a B- in the class.
westothemax is offline  
Old 09-10-2003, 11:07 PM   #3 (permalink)
Upright
 
Thanks for your suggestion. I was able to solve a solution for it, however it's quite lengthy. Here is my solution for anyone who might be interested in it :-)

First, write out the terms of the series:
x = 1*3^1 + 4*3^2 + 9*3^3 +...+ (n-1)^2 * 3^(n-1) + n^2 * 3^n

Then, you multiply that by 3, to shift terms over by one:
3x = 1*3^2 + 4*3^3 +...+ (n-2)^2 * 3^(n-1) + (n-1)^2 * 3^n + n^2 * 3^(n+1)

Now subtract them:
3x-x=2x= -1*3^1 - 3*3^2 - 5*3^3 -...- (-2n+3)*3^(n-1) + (-2n+1)*3^n + n^2 * 3^(n+1)

so now we have what almost looks like a geometric series within a series... we must next form that geo series

multiply by 3 again to shift terms once more:

2x*3=6x= -1*3^2 - 3*3^3 -...- (-2n+5)*3^(n-1) + (-2n+3)*3^n + (-2n+1)*3^(n+1) + n^2 * 3^(n+2)

next, subtract, 6x-2x = 4x to show the geometric series within the series

4x = 1*3^1 + [2*3^2 + 2*3^3 +...+ 2*3^(n-1) + 2*3^n] + (-2n-n^2+1)*3^(n+1) + n^2 * 3^(n+2)

the series contained within the brackets is the geometric series we can sum up. Use the equation [ar^(n+1) - a] / r - 1

2*3^(n+1) - 18 / 3 - 1

Now that we have that, we apply that summation to the remaining 3 terms from the 4x= series....

Thus, 4x = [2*3^(n+1) - 18]/2 + 3 + (-2n-n^2+1)*3^(n+1) + n^2*3^(n+2)

Then get x by itself, so divide by 4:

x = [(2*3^(n+1) - 18)/2 + 3 + (-2n-n^2+1)*3^(n+1) + n^2*3^(n+2)]
/ 4

This is the final answer to find the sums of the first n terms of the series of i^2
* 3^i from i=1 to n.

As you can see, when n=1, the first sum is 3, when n=2, the sum of the first two terms is 39, and when n=3, the sum of the first 3 terms is 282.

It took awhile to derive this, but nonetheless this is a working solution. We have just solved an arithmatic-geometric series. :-D
sieger35 is offline  
Old 09-24-2003, 11:13 PM   #4 (permalink)
Addict
 
btw westothemax, if * is regular multiplication, then you can't break up the sum in such a way.
phukraut is offline  
 

Tags
discrete, math

Thread Tools

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are On
Pingbacks are On
Refbacks are On



All times are GMT -8. The time now is 08:39 AM.

Tilted Forum Project

Powered by vBulletin® Version 3.8.7
Copyright ©2000 - 2017, vBulletin Solutions, Inc.
Search Engine Optimization by vBSEO 3.6.0 PL2
© 2002-2012 Tilted Forum Project

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 288 289 290 291 292 293 294 295 296 297 298 299 300 301 302 303 304 305 306 307 308 309 310 311 312 313 314 315 316 317 318 319 320 321 322 323 324 325 326 327 328 329 330 331 332 333 334 335 336 337 338 339 340 341 342 343 344 345 346 347 348 349 350 351 352 353 354 355 356 357 358 359 360