1) Common ratio = 6r2 = 3r
-1 < (Common ratio) < 1
-1 < 3r < 1
−13<r<13
Please explain. I do not understand what you have just said.
Let a be the first term.a1−−15=16a=16(65)=965
The first term of the series is 96/5
Distribute one candy to each kids first.
5 candies are left, and I can distribute to them in any configuration.
I have 55 = 3125 ways to do so.
(3+5x)7=7∑k=0(7k)⋅37−k⋅(5x)k=7∑k=0(7k)⋅37⋅(53)k⋅xkWhen k=2,Term=(72)⋅37⋅(53)2⋅x2=21⋅243⋅25x2Coefficient of x2=21⋅243⋅25=127575
av+bw+cx=→0(11−12461515)⋅(abc)=(000)(11−10280416)⋅(abc)=(000)(11−1014000)⋅(abc)=(000)Let c=kb+4k=0b=−4ka−4k−k=0a=5ka−bc=5k−(−4k)k=9
k2−81=k2−9k+9k−81=(k2−9k)+(9k−81)=k(k−9)+9(k−9)=(k−9)(k+9)=(k−9)(k+9)
I was careless. I will attempt that again.
(WRONG ANSWER: DELETED)
Denote rn the nth repunit.
S=100∑n=1rn=100∑n=1n−1∑k=010k=110100∑n=1n∑k=110k=110∑1≤k≤n≤10010k=110100∑k=1100∑n=k10k=110100∑k=1(10k(101−k))=110(101(10(10100−1)10−1)−100∑k=1k⋅10k)Let SAG=100∑k=1k⋅10kS=1019(10100−1)−110SAGSAG=1⋅101+2⋅102+⋯+100⋅10100−−−(1)110SAG=1+2⋅101+⋯+100⋅1099−−−(2)(1)−(2):910SAG=100⋅10100−(1+101+102+⋯+1099)=10102−10100−19SAG=101039−10101−1081S=1019(10100−1)−(101029−10100−181)=101009−1019+10100−181=10100−19+10100−90181