The sum 1+(1+3)/2!+(1+3+5)/3!+(1+3+5+7)/4!+… upto ∞ terms, is equal to

Solution

Maths sir · 4 steps

The odd-number sum on top is n², so split n²/n! into two series that each give e.

1. General term. The numerator of the nth term is the sum of the first n odd numbers, which equals n². So the nth term is Tₙ = n²/n!.
2. Break it up. Cancel one n to get n/(n-1)!. Then write n = (n-1) + 1, so for n ≥ 2 the term splits into two pieces.

   n²/n! = n/(n-1)! = 1/(n-2)! + 1/(n-1)!

3. Add each part. The first term (n = 1) is 1 = 1/0!, so summing 1/(n-1)! over all n ≥ 1 gives 1/0! + 1/1! + 1/2! + … = e. Summing 1/(n-2)! over n ≥ 2 also gives e.
4. So. The sum is e + e = 2e, which is option (b).

The idea to keep. For any series with n! in the denominator, rewrite the numerator using n, n(n-1), n(n-1)(n-2) and so on. Each piece cancels against the factorial and leaves a copy of e.

Answer from the NTA final answer key. Solution by PrepWiser.