Answer and Explanation:
Solution: For the general term of each infinite series,
n=1 and the series are:
4/12+1+1/22+1+4/32+1+1/42+1+……..
Let [4/12 +1] + [1/22 + 1] + [4/32+1] + [1/42+1] +………
Sum of infinite series= ∑n=1 [4/n+11 + 1/n] + [1/n+21 + 1/n] + [4/n+31 + 1/n] + [1/n+41 +1/n] +………..
For n=1.