Getting higher power output by using more resistors

Yes, you can use (4) 0.25 W resistors to dissipate 1 W and still remain within each individual resistor's power rating. This can be accomplished a few different ways:

• Place them all in series:
  • In which case you will need to use resistors with 1/4 the resistance that you want overall.
  • e.g. If you want 1 kΩ total, put (4) 250 Ω (240 Ω nearest 5% standard) resistors in series.
• Place them all in parallel:
  • In which case you will need to use ones with 4 times the overall desired resistance
  • e.g. If you want 1 kΩ total, put (4) 4 kΩ (3.9 kΩ nearest std.) resistors in parallel.
• Placing them in a 2x2 array:
  • Where you can use resistors of the same resistance you want overall (2 in parallel gives half, but you place 2 sets of parallel in series, doubling the effective resistance)

In all the mentioned cases, in order to have each resistor dissipate an equal share of power, they all must be equal in value (ohms). This isn't the only way to do it, there are several other combinations you could use with differing values, etc.

Pragmatically, if you're only operating this circuit very intermittently (few seconds at a time), you might be able to get away with a single 1/4 W resistor, especially if this is on a breadboard (be careful not to melt stuff). Higher power resistors often are specified to survive surges of 8-10x their normal power dissipation for several seconds, though the typical 1/4 W thru-hole resistors are carbon film, which is a little less tolerant of this.

Yep, putting them in parallel increases the group power dissipation and lowers group resistance. You still need to calculate power dissipation separately.

$$R1 \parallel R2 = \frac{R_{1}R_{2}}{R_{1}+R_{2}}$$

$$P = IV = I^{2}R = \frac{V^{2}}{R}$$

If you have N resistors in parallel and they are all the same value, power dissipation will be N-times their individual rating and group resistance will be 1/N times their individual rating.