Hello friends can someone tell me how to approach this problem ?? @vijju123 @kaushal101 @mohit_negi @taran_1407 @vivek_1998299 @meooow @john_smith_3 asked 14 Apr, 07:07

Analyze the pattern. Here, it seems that difference between 2 consecutive results is : $N_{i + 1}  N_{i} = 8i + 1 \quad \forall \quad i >= 0 \quad; \quad N_0 = 0$ Use this to come up with formula for $N_i$. I came up with this quadratic formula: $(4i  7)(i  1)$ answered 14 Apr, 09:09
Thanks for your efforts :) Could you explain how did you get that quadratic equation ?
(14 Apr, 12:21)
On using summation on RHS I came up with 4ii+5*i+1 as the answer.
(14 Apr, 12:38)
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@harrypotter0 tbh I was really sleepy when I came up with the equation so I can't remember the exact procedure I went through. To come up with an equation for $n^{th}$ term: Let a function $f(x) = 8x + 1$ Then, $N_1 = f(0) , \quad N_2 = f(0) + f(1)$ and so on So $N_i = \sum_{n=0}^{i1} f(i)$ Use sum of n terms to get a formula: $N_i = 8(0 + 1 + 2 + ... + (i1)) + (1 + 1 + ...(i\quad times)... + 1)$ $N_i = i(4i  3)$ My equation is just for $(i  1)^{th}$ term instead of $i^{th}$ term.
(14 Apr, 19:50)
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@harrypotter0 I forgot to add that your expression gives $(i + 1)^{th}$ term, otherwise its same.. :D
(14 Apr, 22:28)
