Rohit Sharma struck a magnificent double century as India clinched a 57-run victory in the seventh and deciding ODI against Australia.
- Related Content
Sharma's 158-ball demolition took the hosts to an impregnable 383-6 in Bangalore, with Shikhar Dharwan (60) and MS Dhoni (62) providing support with half-centuries.
The tourists made a brave fist of it in reply, James Faulkner (116) cracking a century and proving the last wicket to fall as the Aussies came up short on 326.
A run-rate in excess of seven-and-a-half was always going to be tough on the Aussies as they went in search of a series win with each side having two wins to their name.
But wickets fell in the quest for quick runs and success was unlikely after Brad Haddin was fourth out with 74 on the board in the 16th over.
However, Glenn Maxwell bludgeoned a quick-fire 60 from 22 balls and Shane Watson carried on the good work with a brisk 49 to keep the tourists in with a shout.
Their hopes rested with Faulkner, whose ton came from just 57 balls, but it proved to be in vain.
Earlier Sharma had proved the undoubted star of the show as he produced an innings that broke numerous records.
As well as being the highest ODI knock at the M. Chinnaswamy Stadium, it also contained the most sixes of any ODI innings - a staggering 16, along with 12 fours.
Having taken 112 balls to reach three figures, Sharma needed only a further 42 to reach 200, achieving the feat in the final over of the match, fittingly with a six.
He had shared in an opening stand of 112 with Dhawan, whilst Dhoni produced a typically stylish effort to ensure the innings finished with a bang, smashing one of his two sixes out of the ground.