## Adobe Interview Question for Developer Program Engineers

• 0

Country: India
Interview Type: In-Person

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3
of 3 vote

nth term will be 2n-1.
Sumof(2n-1) for n = 1..n ==> 2(n*(n+1))/2 - n = n^2

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0
of 0 vote

nxn = n(1+(n-1));
take first n odd numbers
1+ 3+ 5...........+2n-1
now take out 1 from every number in above series except from 1.......
n + 2 +4+6+.........+2n-2
=>n+ 2(1+2+3+4.......n-1)
=>n+2 { n(n-1)/2}
=>n+ n(n-1)
=>n^2

hence proved ........... :)

Comment hidden because of low score. Click to expand.
0

How will you prove there are n numbers in sequence....1+3+5+...+2n-1 !!!

Hope question isnt silly?

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0

1 2 3 4 5 6 .................2n are 2n numbers
so they have n numbers of odd and even
now in [1 2n] odd numbers are
1 3 5 7 9............2n-1
so they are n first odd numbers

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0
of 0 vote

let numbers be 1,2,3,......n den odd terms will be 1,3,5...2n-1 so total of n/2 numbers now applyng sum of A.P sum=(no.of terms*(first term+last term))/2 here no of terms is n/2 ,first term is 1 and last is (2*n/2-1)=n so sum=((n/2)*(1+n-1))/2=(n^2)/4 whihc is square of no of terms

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0
of 0 vote

``````Sum of N number = n * (n+1) / 2
sum of 1+2+3+4+5....+2k  = 2k * (2k +1)/2
sum of 1+3+5....+ 2k-1 = 2k^2 + k - 2 * ( sum of 1+2+3+k terms)
sum of 1+3+5... + 2k-1 = 2k^2 + k - 2 * ( k * (k+1)/2)
= k^2``````

Comment hidden because of low score. Click to expand.
0
of 0 vote

Sum of n terms of an AP = (n/2) * (1st term + last term)
Here 1st term is 1 and last term is 2n-1.
So Sum of first n odd numbers is
(n/2) * (1+2n-1) = (n/2) * 2n = n^2

Comment hidden because of low score. Click to expand.
0
of 0 vote

``````mathematical proof
a1 = 1
an = 2* n - 1
so the sequence = 1 + 3 +  5 + 7 + 12 + ...+ 2*n -1
sum = n (a1 + an)/ 2 = n(1 + 2*n - 1) /2= n^2``````

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