Nano Banana 2 image tokens: why Google 4K costs 3.4x 0.5K
Google prices Nano Banana 2 by image token: 747 at 0.5K up to 2,520 at 4K, at $60 per million. Sume quotes a flat price per size. Read 2026-10-03.

Google's Gemini API pricing page (read 2026-10-03) explains the Nano Banana 2 price as a token calculation: standard output is $60 per 1,000,000 tokens, and each resolution uses a fixed number of tokens: 747 at 0.5K, 1,120 at 1K, 1,680 at 2K and 2,520 at 4K. Multiply and you get the per-image figures the page lists.
The calculation
747 tokens at $60 per million is $0.0448, which the page shows as $0.045. 1,120 tokens is $0.0672 ($0.067). 1,680 tokens is $0.1008 ($0.101). 2,520 tokens is $0.1512 ($0.151). The token counts grow 3.4 times from 0.5K to 4K, so the price does too.
| Resolution | Tokens | Google per image | Sume per image | Step up from 0.5K (Google / Sume) |
|---|---|---|---|---|
| 0.5K | 747 | $0.045 | $0.075 | 1.0x / 1.0x |
| 1K | 1,120 | $0.067 | $0.10 | 1.5x / 1.3x |
| 2K | 1,680 | $0.101 | $0.15 | 2.2x / 2.0x |
| 4K | 2,520 | $0.151 | $0.20 | 3.4x / 2.7x |
Sume's curve is flatter
Sume publishes a fixed price per image at each size rather than token pricing, so the quote you see is the quote you pay. Its curve is flatter than Google's: 4K is 2.7 times the 0.5K price on Sume, against 3.4 times on Google. That is because Sume's list basis for the model rises more slowly across sizes ($0.06 to $0.16), and the 1.25 multiplier is applied uniformly.
The practical consequence: the larger the image, the smaller Sume's premium over Google (67 percent at 0.5K, 32 percent at 4K), so 4K is where Sume is closest to Google's price.
A planning rule
Draft at 0.5K and finish only the winners at 2K or 4K. On Google, a 0.5K draft costs a third of a 4K final; on Sume, 37.5 percent. For 20 concepts drafted and 4 finished at 4K, the Sume bill is 20 times $0.075 plus 4 times $0.20, which is $2.30. The same job on Google standard is $0.90 plus $0.604, or $1.50.
Check the arithmetic for your own counts with the snippet.
tokens = {'0.5K': 747, '1K': 1120, '2K': 1680, '4K': 2520}
for size, t in tokens.items():
print(f'{size}: {t} tokens -> ${t / 1_000_000 * 60:.4f} per image at $60/M')
Sources
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Written by Sume