Math, asked by vamankumar1234, 9 months ago

The tens digits of 1! + 2! + 3! + ...... + 49! is?

Answers

Answered by shubham94998

0

Answer:

the tens digits of 1+2+3+4+49

Answered by abhishek123497

0

Answer:

Sn=n/2[2n+(n-1) d]

=49/2[2×(49-1) 1]

=49/2[2(48) ]

=49/2(96)

=49(48)

=2352

at the tens= 5

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