JEE 2023 — Mathematics PYQ
JEE | Mathematics | 2023Let ∑n=0∞(n!)((2n)!)n3((2n)!)+(2n−1)(n!)=ae+eb+c where a,b,c∈Z and e=∑n=0∞n!1. Then a2−b+c is equal to:
Choose the correct answer:
- A.
26
(Correct Answer) - B.
27
- C.
28
- D.
29
26
Explanation
Solution
1. Expression Simplification:
n=0∑∞[n!(2n)!n3(2n)!+n!(2n)!(2n−1)n!]=n=0∑∞n!n3+n=0∑∞(2n)!2n−1
2. Part A calculation (∑n!n3):
n3=n(n−1)(n−2)+3n(n−1)+n
n=0∑∞n!n3=n=3∑∞n!n(n−1)(n−2)+n=2∑∞n!3n(n−1)+n=1∑∞n!n
n=0∑∞n!n3=n=3∑∞(n−3)!1+3n=2∑∞(n−2)!1+n=1∑∞(n−1)!1
=e+3e+e=5e
3. Part B calculation (∑(2n)!2n−1):
n=0∑∞(2n)!2n−1=n=1∑∞(2n)!2n−n=0∑∞(2n)!1
=n=1∑∞(2n−1)!1−n=0∑∞(2n)!1
=(2e−e−1)−(2e+e−1)
=2e−e−1−e−e−1=−e−1=−e1
4. Final Comparison:
Total Sum=5e−e1+0
⇒ae+eb+c=5e+e−1+0
a=5,b=−1,c=0
5. Result:
a2−b+c=52−(−1)+0
25+1=26
Final Answer: 26
Explanation
Solution
1. Expression Simplification:
n=0∑∞[n!(2n)!n3(2n)!+n!(2n)!(2n−1)n!]=n=0∑∞n!n3+n=0∑∞(2n)!2n−1
2. Part A calculation (∑n!n3):
n3=n(n−1)(n−2)+3n(n−1)+n
n=0∑∞n!n3=n=3∑∞n!n(n−1)(n−2)+n=2∑∞n!3n(n−1)+n=1∑∞n!n
n=0∑∞n!n3=n=3∑∞(n−3)!1+3n=2∑∞(n−2)!1+n=1∑∞(n−1)!1
=e+3e+e=5e
3. Part B calculation (∑(2n)!2n−1):
n=0∑∞(2n)!2n−1=n=1∑∞(2n)!2n−n=0∑∞(2n)!1
=n=1∑∞(2n−1)!1−n=0∑∞(2n)!1
=(2e−e−1)−(2e+e−1)
=2e−e−1−e−e−1=−e−1=−e1
4. Final Comparison:
Total Sum=5e−e1+0
⇒ae+eb+c=5e+e−1+0
a=5,b=−1,c=0
5. Result:
a2−b+c=52−(−1)+0
25+1=26
Final Answer: 26