One 4K 1:4 poster vs four stacked 1K squares: $0.20 vs $0.40
A single 1:4 image on Nano Banana 2.1 costs $0.20 at 4K on Sume. Four stacked 1K squares cost $0.40 and do not join. Costs for 0.5K to 4K and when to stack.

Four squares stacked vertically are 1:4 in total, but generating them separately costs $0.40 at 1K, double the $0.20 of one 4K 1:4 image from Nano Banana 2.1 on Sume, and the squares will not join into one scene. One tall call is cheaper and continuous; stacking only makes sense when each square is its own picture.
The comparison
Price is per image for each tier.
| Approach | Images | Price each | Total |
|---|---|---|---|
| One 1:4 at 4K | 1 | $0.20 | $0.20 |
| One 1:4 at 2K | 1 | $0.15 | $0.15 |
| One 1:4 at 1K | 1 | $0.10 | $0.10 |
| Four 1:1 at 2K | 4 | $0.15 | $0.60 |
| Four 1:1 at 1K | 4 | $0.10 | $0.40 |
| Four 1:1 at 0.5K | 4 | $0.075 | $0.30 |
Read the table carefully
The 4K tall image is not four times sharper than four 1K squares; pixel counts depend on the tier and the ratio, and the request does not set exact pixels. What the table shows is the billing: the tall image is one billed generation.
import os, requests
body = {
"model": "google/nano-banana-2.1",
"prompt": "Tall poster with a lighthouse at the top and the sea at the bottom, one continuous scene",
"aspect_ratio": "1:4",
"resolution": "4K",
}
r = requests.post(
"https://api.sume.com/v1/images",
headers={"Authorization": f"Bearer {os.environ['SUME_API_KEY']}"},
json=body, timeout=60,
)
print(r.status_code) # 200 = image body, 202 = job envelope
if r.status_code == 200:
out = r.json()
print(out["data"][0]["url"], out["usage"]["cost"])When to stack squares
- The four panels are different products or different people.
- You need to rerun one panel without paying for the other three.
- Your layout engine wants square files.
- Otherwise ask for 1:4 once. A 4K call can return
202, so handle both response codes.
Sources
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