The sum of 12·21!+22·32!+32·43!+..............∞
5e
3e
7e
2e
Explanation for the correct option
Given: 12·21!+22·32!+32·43!+..............∞
12·21!+22·32!+32·43!+..............∞
nth termTn
⇒Tn=n2(n+1)n!⇒Tn=n(n+1)(n-1)!⇒Tn=(n+2)+2n-1(n-2)!⇒Tn=n+2(n-2)!+2(n-1)(n-2)!⇒Tn=n+4-2(n-2)!+2(n-1)(n-2)!⇒Tn=n-2(n-2)(n-3)!+4(n-2)!+2(n-1)!⇒Tn=1(n-3)!+4(n-2)!+2(n-1)!
put n=3 and formatting T3 in the following form
⇒T3=1(3-3)!+4(3-2)!+2(3-1)!⇒T3=10!+41!+22!
similarly,
T4=11+23!+42!
T5=12!+24!+43!
T1=12·21!⇒21!T2=22·32!⇒6⇒20!+40!
So,
T1+T2+T3+.........=21!+20!+40!+10!+41!+22!+................T1+T2+T3+......=10!+11!+12!+.........+210!+11!+12!+.........1+410!+11!+12!+.........---------(1)
the expansion for ex=1+x1!+x22!+x33!+.........
put x=1
e1=1+11!+122!+133!+.........e=1+11!+12!+13!+.........
put the value in (1)
T1+T2+T3+.........=e+2e+4eT1+T2+T3+.........=7e
Hence, option C is correct.
Find the sum of the following series to infinity :
(i) 1−13+132−133+134+....∞
(ii) 8+4√2+4+....∞
(iii) 25+352+253+354+.....∞.
(iv) 10 - 9 + 8.1 - 7.29 + ...... ∞
(v) 13+152+133+154+135+156+.....∞