Induction Question (1 Viewer)

JaxsenW · Mar 1, 2018

Could someone please provide the working out for Question 13a)
*Edit: preferably using Series notation
Thanks

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integral95 · Mar 1, 2018

$n = 1 \Rightarrow \\ \\ LHS = 1*1! = 1 \ \ \ RHS = (1+1)! - 1 = 2 - 1 = 1$

True for n = 1

$$ Assume it's true for $ \ \ n = k \ \ \\ \\ \sum_{r=1}^{k} r*r! = (k+1)! -1 \\ \\ $Then we prove it's true for $ \ \ n = k+1 \ \ \Rightarrow \\ \\ \sum_{r=1}^{k+1} r*r! = (k+2)! -1$

$LHS = \sum_{r=1}^{k+1} r*r! \\ \\ = \sum_{r=1}^{k} r*r! + (k+1)(k+1)! \\ \\ = (k+1)! -1 + (k+1)(k+1)! \\ \\ = (k+1)!(1+k+1) - 1 \\ \\ = (k+1)!(k+2) - 1 \\ \\ = (k+2)! - 1 = RHS$

All done

JaxsenW · Mar 1, 2018

integral95 said:
$n = 1 \Rightarrow \\ \\ LHS = 1*1! = 1 \ \ \ RHS = (1+1)! - 1 = 2 - 1 = 1$

True for n = 1

$$ Assume it's true for $ \ \ n = k \ \ \\ \\ \sum_{r=1}^{k} r*r! = (k+1)! -1 \\ \\ $Then we prove it's true for $ \ \ n = k+1 \ \ \Rightarrow \\ \\ \sum_{r=1}^{k+1} r*r! = (k+2)! -1$

$LHS = \sum_{r=1}^{k+1} r*r! \\ \\ = \sum_{r=1}^{k} r*r! + (k+1)(k+1)! \\ \\ = (k+1)! -1 + (k+1)(k+1)! \\ \\ = (k+1)!(1+k+1) - 1 \\ \\ = (k+1)!(k+2) - 1 \\ \\ = (k+2)! - 1 = RHS$

All done

Thankyou! I see where I went wrong now

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Induction Question (1 Viewer)

JaxsenW

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integral95

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JaxsenW

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