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>
Answer:
b
Step-by-step explanation:
hgvfigjl
Answer:
x = -2
Step-by-step explanation:
1) add 2 to both sides to eliminate -2.
5x = -10
2) divide both sides by 5.
x = -2
I think your answer is correct if not the last one because there is an outlier and with process of elimination all the others ones that say there isn't are incorrect and unless that answer you have bubbled is incorrect than your last resort is the One cluster One outlier but i honestly dont see a true cluster :)
