1 - 1 -2 - 2 -3 - 3 - - - n - n - n -1 -1 for all N - N-

Consider equation 1 × 1! + 2 × 2! + 3 × 3! + ..... + n × n! Lets take (n + 1)! n! = n! (n+ 1 1) = n× n! Now substitute n = 1, 2, 3, 4, .... n in above equat ...

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Standard XII

Mathematics

Proof by mathematical induction

1 × 1 !+2 × 2...

Question

$1 \times 1! + 2 \times 2! + 3 \times 3! + . . . . . + n \times n! = (n + 1)! - 1$ for all N $ϵ$ N.

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Solution

Consider equation

$1 \times 1! + 2 \times 2! + 3 \times 3! + . . . . . + n \times n!$

Lets take $(n + 1)! - n! = n! (n + 1 - 1)$

$= n \times n!$

Now substitute $n = 1, 2, 3, 4, . . . . n$ in above equation we get

$2! - 1! = 1 \times 1!$

$3! - 2! = 2 \times 2!$

$4! - 3! = 3 \times 3!$

..........

$(n + 1)! - n! = n \times n!$

Adding all the above terms gives

$1 \times 1! + 2 \times 2! + 3 \times 3! + . . . . . . . n \times n! = 2! - 1! + 3! - 2! + 4! - 3! . . . . + (n + 1)! - n!$

$1 \times 1! + 2 \times 2! + 3 \times 3! + . . . . . . n \times n! = (n + 1)! - 1$

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1×1!+2×2!+3×3!+.....+n×n!=(n+1)!−1 for all N ϵ N.

$1 \times 1! + 2 \times 2! + 3 \times 3! + . . . . . + n \times n! = (n + 1)! - 1$ for all N $ϵ$ N.