1hey for the 7th question i m getting answer for (n2+n-1)/(n+1) but not for (n2+n+1)/(n+1) .please check ur question
Please give solutions for these:-
1) If a2+b2+c2=4, x2+y2+z2=9 then what is the maximum value of ax+by+cz ?
2) Sum of the series upto 11 terms : 1.2.3.4+2.3.4.5+..........+n(n+1)(n+2)(n+3)
3) Prove 1.3.5.7......upto 100 terms<100100
4) The (26-1)th term of a HP is 213+27 and the (26+1)th term is 213-27. Then 27th term is ?
5) If S=1! + 2 X 2! + 3 X 3!+......+ 10 X 10!, then S+2 when divided by 11! leaves a remainder of ?
6) If a,b,c are positive integers satisfying [a/(b+c)] + [b/(c+a)] + [c/(a+b)] then what is the value of abc+(1/abc) ?
7) Σ(n=0 to ∞) (n2+n+1)/(n+1)! = ?
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7 Answers
1hey check ur seventyh question , there might be some error
1
23n2+n+1(n+1)!=n(n+1)+1(n+1)!
=nn!+1(n+1)!
ex= Σ xnn!
differential both sides wrt x
ex= Σ nxn-1n!
put x = 1
e = Σ nn!
also ex = Σ(n=-1 to ∞) xn+1(n+1)!
put x = 1
e = Σ(n=-1 to ∞) 1(n+1)!
i.e
e= 1 +Σ 1(n+1)!
thus S = e + e -1 = 2e -1
23Tn=n(n+1)(n+2)(n+3)
let Vn= n(n+1)(n+2)(n+3)(n+4)
Vn+1=(n+1)(n+2)(n+3)(n+4)(n+5)
Vn+1 - Vn= 5(n+1)(n+2)(n+3)(n+4)=5Tn+1
i.e
Vn - Vn-1=5Tn
hence Σ Tn =Vn -V05
713.)
We have GM < AM
Therefore,
100√ (1.3.5.7....) < (1+3+5+7+....) / 100
=> 100√ (1.3.5.7....) < 100
=> Raising to Power 100
=> (1.3.5.7.... upto 100 terms) < 100100
1for the first one applying caushy schwartz inequality we get it as 6....
