The correct option is C 2
S1=12+16+112+120+⋯∞=∞∑n=11n(n+1)=∞∑n=1(1n−1n+1)=limn→∞(11−12)+(12−13)+(13−14)+⋯+(1n−1n+1)=limn→∞(1−1n+1)∴S1=1−limn→∞1n+1=1
S2=13+152+133+154+135+156+⋯∞=(13+133+135+⋯∞)+(152+154+156+⋯∞)=131−132+1521−152=38+124=512∴S1−2S2=1−2×512=212∴a=2