The expression (1+x²/2! + x⁴/4! + x⁶/6! +…)² will be represented in ascending power of x as

(1) 1+ 2²x²/2! + 2⁴x⁴/4! + …

(2) 1+(2x)²/2! + 2²x⁴/4! + …

(3) 1+(2x)²/2.2! + 2x⁴/4! + …

(4) 1+(2x)²/2.2! + (2x)⁴/2.4! +…

Solution:

We know e^x = 1 + x/1! + x²/2! + x³/3! +…

e^-x = 1 – x/1! + x²/2! – x³/3! +…

(e^x + e^-x) = 2 (1+x²/2! + x⁴/4! +…)

(1+x²/2! + x⁴/4! +…) = (e^x + e^-x) /2

(1+x²/2! + x⁴/4! + x⁶/6! +…)² = ((e^x+e^-x)/2)²

= (1/4)(e^2x + e^-2x + 2)

= (1/4)[ 4 + 2{(2x)²/2! + (2x)⁴/4! + …}]

= 1 + (2x)²/2.2! + (2x)⁴/2.4! + ….

Hence option (4) is the answer.