which series is it 2+6+12 and so on to n
and please unnecessary chatting otherwise I report you
Answers
Answer:
1^1+1=2
2^2+2=6
3^2+3=12
4^2+4=20
5^2+5=30
nth term=n^2+n
Next number=6^2 +6=36+6=42
II method:
1*(1+1)=2
2*3=6
3*4=12
4*5=20
5*6=30
nth term=n*(n+1)
next no=6th term= 6*(6+1)=6*7=42
EDIT 1:
III METHOD:
0+2*1=2
2+2*2=6
6+2*3=12
12+2*4=20
20+2*5=30
Thus nth term
Tn=Tn-1+2*n Or Tn-Tn-1=2*n
So we can write
T2-T1=2*2
T3-T2=2*3
T4-T3=2*4
………………..
………………..
Tn-1-Tn-2=2*(n-1)
Tn-Tn-1=2*n
————————————————
Adding all we get
Tn-T1=2*(2+3+4+………………………….+n)
Tn=2*(1+2+3+4+………………………….+n)
=2*n(n+1)/2=n(n+1)
Tn=n(n+1)
which is same as above
Thus T6=T5+2*6=30+12=42
METHOD IV:
Tn=(n^3–1) /(n-1) - 1
Thus T3=(27–1)/(3–1)-1=26/2–1=13–1=12
T4=(64–1)/(4–1) -1 =63/3–1=21–1=20
T5=(125–1) / ( 5–1)-1=124/4–1=31–1=30
Next term=T6=(6^3–1)/(6–1) -1=(216–6)/(6–1) -1=215/5–1=43–1=42
EDIT 2:
METHOD V:
nth term=(n+1)(n-1) +n+1
T1=0*2+2=2
T2=1*3+3=6
T3=2*4+4=12
T4=3*5+5=20
T5=4*6+6=24+6=30
Next term T6=5*7+7=35+7=42
EDIT 3:
METHOD VI:
nth term=(n-1)^2 +3n-1
T4=(4–1)^2+3*4–1=9+12–1=20
T5=(5–1)^2 +3*5–1=16+15–1=30
Next term T6=(6–1)^2 +3*6–1=25+18–1=42
Answer:
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