1any hint ?
sir , how to solve non linear recurrence , like this one , i mean for linear u told one techonique
A recurrence satisfies a_{n+1} = a_n + \sqrt{a_n+a_{n+1}}
a1 =1
Obtain an explicit formula for an i.e. find a function f(n) such that an = f(n)
edited: first term provided now, sry.
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10 Answers
1
1i found first few terms of series
thjey came out to be
1,3,6,10
so
an-an-1=ln+c=linear
so
an=quadratic in n
an=ln2+mn+c , now put
a1=1
a2=3
a3=6
we l=1,m=1 c=0
hence an=n(n+1)/2
not a strong proof though
11f(1)=1
f(2)=f(1 )+2=3
f(3)=f(2 )+3=6
Induction gives f(n)=f(n-1)+n =f(n-2)+(n-1)+n=f(n-3)+(n-2)+(n-1)+n
Continuing like this, we finally have f(n)=f(1)+2+3+4+...+n=\frac{n(n+1)}{2}.
Happy xyz?
49a_{n+1}=a_{n}+\sqrt{a_{n+1}+a_{n}}
\Rightarrow (\sqrt{a_{n+1}+a_{n}})^2-\sqrt{a_{n+1}+a_{n}}-2a_{n}=0
\Rightarrow \sqrt{a_{n+1}+a_{n}}=\frac{1\pm \sqrt{1+8a_{n}}}{2}
\Rightarrow a_{n+1}+a_{n}=\frac{1+1+8a_{n}\pm 2\sqrt{1+8a_{n}}}{4}
\Rightarrow a_{n+1}=\frac{2+4a_{n}\pm 2\sqrt{1+8a_{n}}}{4}
\Rightarrow a_{n+1}=a_{n}+\frac{2\pm 2\sqrt{1+8a_{n}}}{4}
{\color{red}\Rightarrow a_{n+1}=a_{n}+\frac{1\pm \sqrt{1+8a_{n}}}{2}}
does that help much????[17][17][17]
1TRY THIS QUESTION:
A recurrence satisfies
a_{n+1}= a_{n}+\sqrt{1+a_{n}}
Obtain an explicit formula for a_{n} i.e. find a function f(n) such that a_{n}=F(n)
(given that a_{0}=0)
1hey soumik mind ur attitude[16]
after assuming an=n(n+1)/2
directly put it in recurrence relation ,u will get it true for an+1