To start, assume all cells can take all values

10 cage in r6/7 must be 1,2,3,4 - eliminate 5/6/7/8/9

15 cage in r1/2/3 must be 1,2,3,4,5 - eliminate 6/7/8/9

21 cage in r3/4 must be 1,2,3,4,5,6 - eliminate 7/8/9

32 cage in c1/2 cannot contain a 1 - eliminate it

11 cage in c9 cannot contain a 1 - eliminate it

21 cage in c9 cannot contain 1/2/3 - eliminate them

19 cage in c1 cannot contain a 1 - eliminate it

26 cage in r7 cannot contain a 1 - eliminate it

7 cage in r9 cannot contain 7/8/9 - eliminate them

12 cage in r8 cannot contain 1/2 - eliminate them

7,12 and 10 cages in r9 sum to 29, leaving the remaining two cells to make 16. This must be 7,9, so eliminate 1/2/3/4/5/6/8 from these two cells

Naked pair of 7/9 in r9

Remaining two cells of 28 cage sum to 12 - must be 4,8, so eliminate 1/2/3/5/6/7/9

Naked pair of 4/8 in r8

10 cage in r9 cannot contain 5 - eliminate it

10 cage in r9 cannot be 3,7 or 1,9 - eliminate 1/3

12 cage in r8 cannot contain 6 - eliminate it

10 cage in r8 cannot contain 5, else it would be 2,3,5, invalidating the 12 cage in r8.

10 cage in r8 must contain a 1 - eliminate 1 from other cells in r8

7,28,10,12,12,10 cages in r8/9 sum to 79, leaving 11. The possibilities for 11 are 2&9, 5&6 - eliminate 3/7

11,21 cages in r9 sum to 32, leaving 13 in four cells. Eliminate 8/9 from these four cells

Can eliminate 2 from r9c8 due to the 10 cage, and 3 from r8c8 due to the 12 cage

Going back to that 13 in four cells, no way of making it if there's a 4 in r6 or r7, so eliminate

Similarly, can remove 3 from r8c9, and 6 from r9c9

Then remove 9 from r8c8, 4 from r9c8, due to the 12 and 10

Naked pair 5/7 in r8 and box 9

10 cage in r8 must now be 1,3,6 - eliminate 2 from this cage, and 6 from other cells in the row

Naked pair 2/9 in box 7

Eliminate 5 from the 7 cage in r9

5 in r9 constrained to the 12 cage, so 12 cage cannot contain an 8

Hidden single 8 in r9, and also fill in 2 in r9c9

Naked pair 1/3 in c9 and in 10 cage in c8/9

Naked singles 4,2 in that 10 cage

To make the bottom four cells of c9 sum to 13, r8c9 must be 7 - then r8c8 is 5.

Hidden single 9 in box 9

6 in box 9 constrained to c7 - eliminate 6 from rest of c7

11 cage in c9 must be 5,6 - eliminate other values

Naked pair 5/6 in c9 and box 3

If r9c5 or r9c6 is 3, cannot complete the 12 cage, so eliminate it from them

5 in r9 constrained to box 8 - eliminate other 5s from box 8

Hidden single 9 in box 8 - also fill in the 7 in r9c3

r7c4,r7c5,r7c6,r8c5,r8c6,r8c7,r9c5,r9c6,r9c7,r9c4 sum to 48 - this means r8c7 + r9c7 is 3 more than r8c4. As This sum cannot be 11, r8c4 cannot be 8.

Naked single 4 in r8c4, also fill in the 8 in r8c3

Now know that r8c7 + r9c7 is 7, and so r7c9 is 3 - also fill in the 1 in r6c9

Naked pair 1/6 in c7

Naked triple 1/5/6 in r7 and box 7

19 cage in c1 can only be 4,6,9 - fill in these digits

Naked singles 2,3 in box 7

Naked single 4 in r9c2

Cages 32+17+19+7+24 in c1/2/3 sum to 99 - this means r6c3 + r7c3 is 9 more than r1c2. Hence, r1c2 cannot be a 3

Can see that r6c2 + r6c3 must be 16 - fill in the 7 and 9

6 in r6 constrained to box 5 - eliminate other 6s from the box

One of r3c1,r4c1,r4c1 must be an 8. If not, we cannot fill the 17 cage - so eliminate 8 from other cells in c1

Looking at the 32 cage shows r2c2,r3c2 and r4c2 cannot be 5, or we cannot complete the cage - eliminate them

The ALS r3c1,r4c1,r5c1 excludes either 2 or 5 - this then leads to 3,6 as the only possibilities for r5c2

9 in r4 constrained to box 6 - eliminate other 9s

6 in c3 constrained to box 4 - eliminate other 6s

Naked single 3 in r5c2

Naked single 8 in r4c2

The ALS r3c1,r4c1,r5c1 now can't contain 2, to make the 17 cage consistent

Naked pair 1/5 in c2 and box 4

Naked pair 2/6 in c3

Naked single 8 in r3c1

Naked pair 4/9 in c9

Naked single 8 in c9

The 2 in the 15 cage can now only go in r1c4 - put it there

Naked single 7 in r1c1

Naked single 2 in r2c1

Hidden single 2 in box 3

31 cage in c7/8 contains a 2 and has no 8 - the only combination possible is 2,3,4,6,7,9 - eliminate 1s and 5s

Hidden single 5 in box 6

Naked triple 3/6/8 in box 5

Naked pair 1/5 in r4

Naked pair 2/4 in r4 and box 5

Naked singles 6,9 in r4

Naked single 4 in c9

Naked single 2 in r5

Naked pair 3/7 in box 6

Naked singles 4,6 in box 6

31 cage must have a 9 in r3c8

Naked single 6 in r3c2

Naked triple 1/3/5 in r3

Naked single 9 in r2c2

Naked single 7 in r3c6

34 cage and 11 cage in (ish) box 3 add to 45, meaning r1c5 + r1c6 is 15. The only posibility is 6/9, so eliminate all others

Naked pair 6/9 in r1 and box 2

Naked single 5 in r1c9

Naked single 1 in r1c2

Naked single 5 in r7c2

Naked single 1 in r7c3

Naked single 3 in r1c8

Naked single 4 in r1c3

Naked single 6 in r2c9

Naked single 7 in r4c8

Naked single 1 in r2c8

Naked single 3 in r4c7

The only combination for the 24 cage in r2 (ish) is 4,5,7,8 - eliminate 3 from cells

Naked triple 4/5/8 in r2

Naked singles in r2

Naked singles in boxes 1/3

Naked pair 1/3 in 21 cage

Naked single 5 in r4c4

Naked single 1 in r4c1

Naked single 5 in r5c1

Naked single 8 in r2c4

Naked single 7 in r7c4

Naked single 1 in r5c4

Naked single 3 in r3c4

Naked single 1 in r3c5

Naked single 9 in r5c6

Naked single 7 in r5c5

r6c6 must be 3 to finish the 24 cage

Naked single 6 in r6c4

Naked single 8 in r6c5

Naked single 2 in r7c5

Naked single 8 in r7c6

Naked single 4 in r4c5

Naked single 2 in r4c6

Naked single 5 in r2c5

Naked single 4 in r2c6

Naked single 6 in r9c5

Naked single 1 in r9c7

Naked single 5 in r9c6

Naked single 6 in r8c7

Naked single 1 in r8c6

Naked single 3 in r8c5

Naked single 9 in r1c5

Naked single 6 in r1c6

Done!