Sequences and Series Part 2

Given,
an=an−1+2
⇒an−an−1=2
∴ In this series between any two consecutives terms difference is 2 . So this is an A.P. with common difference 2.
Also given a1=1
∴ Series is =1,3,5,7......
∴an=1+(n−1)2=2n−1
Also bn=an+bn−1
When n=2 then
b2−b1=a2=3
.⇒b2−1=3‌ [Given ‌b1=1]
⇒b2=4
When n=3 then
b3−b2=a3
⇒b3−4=5
⇒b3=9
∴ Series is =1,4,9......
=12,22,32......n2
∴bn=n2
Now,

15

∑

n=1

(an.bn)
=

15

∑

n=1

[(2n−1)n2]
=

15

∑

n=1

2n3−

15

∑

n=1

n2
=2(13+23+...153)−(12+22+...152)
=2×(‌

15×16

2

)2−(‌

15(16)×31

6

)
=27560