Question

last digit of 1^5+2^5+3^5.........99^5 is​

Answer 1

Step-by-step explanation:

$\huge\underline\mathfrak\color{blue} Answer$

We know that,

( 15+25+35+……..+n5 )

= [n(n+1)]2(2n2+2n–1)12

So, for ( 15+25+35+……..+995 )

The answer will be,

161708332500

So, the last digit is zero.

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

∑nk=1kp=np+1p+1+12np+∑pk=2Bkk!pk−1–––––np−k+1

where, pk−1–––––=p!(p−k+1)! is called a falling factorial and Bk are the Bernoulli numbers.

Using that formula we can deduce any specific formula for power sum, as it is given below:

Σn0=n

Σn1=n(n+1)2=12(n2+n)

Σn2=n(n+1)(2n+1)6=16(2n3+3n2+n)

Σn3=[n(n+1)2]2=14(n4+2n3+n2)

Σn4=n(n+1)(2n+1)(3n2+3n−1)30=130(6n5+15n4+10n3−n)

Σn5=[n(n+1)]2(2n2+2n–1)12=112(2n6+6n5+5n4−n2)

Σn6=n(n+1)(2n+1)(3n4+6n3−3n+1)42=142(6n7+21n6+21n5−7n3+n)

Σn7=n2(n+1)2(3n4+6n3−n2−4n+2)24=124(3n8+12n7+14n6−7n4+2n2)

Σn8=n(n+1)(2n+1)(5n6+15n5+5n4−15n3−n2+9n−3)90=190(10n9+45n8+60n7−42n5+20n3−3n)

Σn9=n2(n+1)2(n2+n−1)(2n4+4n3−n2−3n+3)20=120(2n10+10n9+15n8−14n6+10n4−3n2)

Σn10=n(n+1)(2n+1)(n2+n−1)(3n6+9n5+2n4−11n3+3n2+10n−5)66=166(6n11+33n10+55n9−66n7+66n5−33n3+5n)