Sequence induction proof (1 Viewer)

MoonMonster196884 · Mar 8, 2024

A sequence is defined by $b_{1}=1$ and $b_{n+1}=b_{n}\left(b_{n}+1\right)$ . Prove that for each n: $b_{n+1}=1+\sum_{r=1}^{n}b_{r}^{2}$

Drongoski · Mar 8, 2024

In outline as LaTeXing is so tedious.

$\therefore b_2 = b_{1+1} = b_1(b_1 + 1) = 1(1+1) = 2\\ \\ \therefore b_2 = 2 = 1 + \sum ^1 _{r=1} b_r ^2 = 1 + 1^2$

Therefore true for k = 1. Now assume true for n = k (k >= 1)

$\therefore b_{k+1} = 1 + \sum ^k _{r=1} b_r ^2 \\ \\ b_{k+2} = b_{k +1}(b_{k + 1} + 1) = b_{k+1} ^2 + b_{k+1} = 1 + \sum ^ k _{r=1} b_r ^2 + b_{k+1} ^2 = 1 + \sum ^{k+1}_{r=1} b_r ^2 = b_{(k+1) + 1}$

i.e. if true for n=k true also for n=k+1

. . . blah ... blah ... blah . . .

MoonMonster196884 · Mar 8, 2024

it's very easy i did it.
Assume $b_{k}=1+\sum_{r=1}^{k-1}b_{r}^{2}$
RTP that for k+1
We know $b_{k+1}=b_{k}\left(b_{k}+1\right)$
=$b_{k}^{2}+b_{k}=b_{k}^{2}+1+\sum_{r=1}^{k-1}b_{r}^{2}$ by assumption
=what we were RTP

Sequence induction proof (1 Viewer)

MoonMonster196884

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Drongoski

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MoonMonster196884

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