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If pm stands for mpm then 1+1p1+2p2+3p3+.....+npn equal to

1!*2!*3!*4!*5!*...*n! as npn=n!
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sorry 0!+1!+2!+...+n! (i have written 1 as 0!)
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1! + 2*2! + 3*3! + ... + n*n! = (n+1)! - 1�

by induction :�

for n = 1, 1! = 1 and (1+1)! - 1 = 2 - 1 = 1 : it's true�

supposing it's true for n, we proove it's true for n+1 :�

1! + 2*2! + ... + n*n! + (n+1)*(n+1)! = (n+1)! - 1 + (n+1)*(n+1)!�
= (n+1)! (1 + n+1) - 1�
= (n+2)(n+1)! - 1�
=(n+2)! - 1�

it's true for n+1, then it's true for all n�1! + 2*2! + 3*3! + ... + n*n! = (n+1)! - 1�

by induction :�

for n = 1, 1! = 1 and (1+1)! - 1 = 2 - 1 = 1 : it's true�

supposing it's true for n, we proove it's true for n+1 :�

1! + 2*2! + ... + n*n! + (n+1)*(n+1)! = (n+1)! - 1 + (n+1)*(n+1)!�
= (n+1)! (1 + n+1) - 1�
= (n+2)(n+1)! - 1�
=(n+2)! - 1�

it's true for n+1, then it's true for all n�1! + 2*2! + 3*3! + ... + n*n! = (n+1)! - 1�

by induction :�

for n = 1, 1! = 1 and (1+1)! - 1 = 2 - 1 = 1 : it's true�

supposing it's true for n, we proove it's true for n+1 :�

1! + 2*2! + ... + n*n! + (n+1)*(n+1)! = (n+1)! - 1 + (n+1)*(n+1)!�
= (n+1)! (1 + n+1) - 1�
= (n+2)(n+1)! - 1�
=(n+2)! - 1�

it's true for n+1, then it's true for all n�
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I DN'T KNOW
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IDNK
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mPm = m! Now the series will be 1+1!+ 2(2!)+ 3(3!)+ 4(4!) .+.....+ n(n!) Now ,we know that r(r!) = (r+1)!-r! The series will change to : 1+ (2!-1!)+ (3!-2!)+ (4!-3!)+ (..)....+((n+1)!-n!) All the terms gets cancelled out and we are left with (n+1)! So This series is for ==>> (n+1)!
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If two objects A and B apply force on each other. Then, Fab = - Fba

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