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Question Number 200718 by hardmath last updated on 22/Nov/23

3. 1+3+5+7+...+(2n+1) = ?

3.1+3+5+7+...+(2n+1)=?

Answered by deleteduser1 last updated on 22/Nov/23

2+4+..+2n=2(1+2+3+...+n)=n^2 +n 1+2+3+...+2n+1=(2n+1)(n+1) 1+3+5+...+2n+1=(n+1)(2n+1−n)=(n+1)^2

2+4+..+2n=2(1+2+3+...+n)=n2+n1+2+3+...+2n+1=(2n+1)(n+1)1+3+5+...+2n+1=(n+1)(2n+1−n)=(n+1)2

Answered by mr W last updated on 22/Nov/23

S=Σ_(k=0) ^n (2k+1) =2Σ_(k=0) ^n k+n+1 =2×((n(n+1))/2)+n+1 =(n+1)^2

S=∑nk=0(2k+1)=2∑nk=0k+n+1=2×n(n+1)2+n+1=(n+1)2

Answered by Mathspace last updated on 23/Nov/23

1+3+5+7+....+2n+1 2n+1+2n−1+2n−3+.... Σ=2n+2+2n+2+2n+2+...(n+1)× =2(n+1)^2 =2s ⇒ s=(n+1)^2

1+3+5+7+....+2n+12n+1+2n−1+2n−3+....Σ=2n+2+2n+2+2n+2+...(n+1)×=2(n+1)2=2s⇒s=(n+1)2

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