Question

what is the unit digit of 1^5+2^5+3^5+.......+98^5​

Answer 1

Answer:

If we do n5n5 the last digit of the result always comes the same as the last digit of n.

Now, if we add ( 15+25+35+……..+99515+25+35+……..+995 )

The last digit will come as the last digit of the addition (1+2+3+…..+99).

Now,

The last digit of the addition (1+2+3+…..+99)

= The last digit of 99×(99+1)299×(99+1)2

= The last digit of 99×100299×1002

= 0

So, the last digit of the addition,

( 15+25+35+……..+99515+25+35+……..+995 ) will be Zero.

Process 2:

We know that,

( 15+25+35+……..+n515+25+35+……..+n5 )

= [n(n+1)]2(2n2+2n–1)12[n(n+1)]2(2n2+2n–1)12

So, for ( 15+25+35+……..+99515+25+35+……..+995 )

The answer will be,

161708332500

So, the last digit is zero.

P.S: We know that 1a+2a+3a+……..+na1a+2a+3a+……..+na is written mathematically as ΣnaΣna . The general formula for the power sum is known as Faulhaber's formula (also known as Bernoulli's formula):

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