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kvasek [131]
3 years ago
7

WILL GIVE BEST RESPONSE

Mathematics
1 answer:
tino4ka555 [31]3 years ago
4 0
We will use demonstration of recurrences<span>1) for n=1, 10= 5*1(1+1)=5*2=10, it is just
2) assume that the equation </span>10 + 30 + 60 + ... + 10n = 5n(n + 1) is true, <span>for all positive integers n>=1
</span>3) let's show that the equation<span> is also true for n+1, n>=1
</span><span>10 + 30 + 60 + ... + 10(n+1) = 5(n+1)(n + 2)
</span>let be N=n+1, N is integer because of n+1, so we have
<span>10 + 30 + 60 + ... + 10N = 5N(N+1), it is true according 2)
</span>so the equation<span> is also true for n+1, 
</span>finally, 10 + 30 + 60 + ... + 10n = 5n(n + 1) is always true for all positive integers n.
<span>
</span>
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