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The Brainliest Answer!
2014-04-28T11:52:45+05:30
We'll use: e = Σ [n=0→∞] 1/n! 

S = Σ [n=1→∞] (4n² - 1)/n! 
S = Σ [n=1→∞] (4.n(n-1) + 4n -1)/n! 
S = 4.Σ [n=1→∞] n(n-1)/n! + 4.Σ [n=1→∞] n/n! - Σ [n=1→∞] 1/n! 

Let's trim the initial number of the sum, using n(n-1)=0 for n=1 and 1/n! = 1 for n=0 
S = 4.Σ [n=2→∞] n(n-1)/n! + 4.Σ [n=1→∞] n/n! - Σ [n=0→∞] 1/n! + 1 

We may now simplify and start counting from zero: 
S = 4.Σ [n=2→∞] 1/(n-2)! + 4.Σ [n=1→∞] 1/(n-1)! - Σ [n=0→∞] 1/n! +1 
S = 4.Σ [p=0→∞] 1/p! + 4.Σ [q=0→∞] 1/q! - Σ [n=0→∞] 1/n! +1 
S = 7.Σ [n=0→∞] 1/n! + 1 
S = 7e + 1
1 5 1
thanq
wlcm
i have given another question can you plz help with that also?
let me chk it now for sure.