Question

1+2+3+----------------------+n
___________________. =. ?
1+3+5+ -----------------+(2n-1)​

Answer 1

1+2+3+n=6+n

1+3+5+___= 2n-1

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Answer 2

Answer:

So,

let P(k):1+3+5…+2k−1=k2P(k):1+3+5…+2k−1=k2

P(1):1=12P(1):1=12 (the given equation is true for n = 1)

let it be true for kth value, ie.

let P(k):1+3+5+...+2k−1=k2P(k):1+3+5+...+2k−1=k2

then 1+3+5+…+2k−1+2(k+1)−1=k2+2(k+1)−11+3+5+…+2k−1+2(k+1)−1=k2+2(k+1)−1

=>1+3+5+...+2k−1+2k+1=k2+2k+1=>1+3+5+...+2k−1+2k+1=k2+2k+1

the RHS of the equation gives (k+1)^2

=>1+3+5+…+2k−1+2k+1=(k+1)2:P(k+1)=>1+3+5+…+2k−1+2k+1=(k+1)2:P(k+1)

so we proved that P(1) is true and P(k+1) is true when P(k) is true.

hence 1+3+5+…+(2n−1)=n21+3+5+…+(2n−1)=n2

kindly note that this in an example

Step-by-step explanation:

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