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a-n-1-a-n-n-n-a-1-1-a-n-n-N-

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Question Number 2644 by 123456 last updated on 24/Nov/15
a_(n+1) =(a_n /n)+n a_1 =1 a_n =??? n∈N^∗
an+1=ann+na1=1an=???n∈N∗
Commented by RasheedAhmad last updated on 24/Nov/15
a_(n+1) =(a_n /n)+n a_1 =1 a_n =??? n∈N^∗ −−−−−−−−−− Let n→n−1 a_(n+1) =(a_n /n)+n ⇒a_n =(a_(n−1) /(n−1))+n−1
an+1=ann+na1=1an=???n∈N∗−−−−−−−−−−Letn→n−1an+1=ann+n⇒an=an−1n−1+n−1
Answered by RasheedAhmad last updated on 24/Nov/15
a_(n+1) =(a_n /n)+n a_1 =1 a_n =??? n∈N^∗ −−−−−−−−−− n→ n−1 a_(n+1) =(a_n /n)+n ⇒a_n =(a_(n−1) /(n−1))+n−1 a_1 =1 [given] a_2 =a_1 +1=2 a_3 =(a_2 /2)+2=3 ..... .... a_n =n This can be proved using induction.
an+1=ann+na1=1an=???n∈N∗−−−−−−−−−−n→n−1an+1=ann+n⇒an=an−1n−1+n−1a1=1[given]a2=a1+1=2a3=a22+2=3…..….an=nThiscanbeprovedusinginduction.

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