If - r -0 n r 2- r -1 r -2016 - 2016 - then n equals to-

∑_r=0^n (r^2+r+1)r!=2016 × 2016! ∑_r=0^n (r^2+r+1)r!=(n+1)!. (n+1) ∴ (n+1)(n+1)! = 2016 × 2016! ∴ n+1 = 2016 ∴ n=2015 ...

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Sum of Coefficients of All Terms

If ∑ r =0 n r...

Question

If $\sum_{r = 0}^{n} (r^{2} + r + 1) r! = 2016 \times 2016!$ then n equals to,

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T_{r} =^{2016} C_{r} x^{2016 - r}

r = 0, 1,, . . . .2016

(T_{0} - T_{2} + T_{4} . . . . + T_{2016})^{2} + (T_{1} - T_{3} + T_{5} . . . . T_{2015})^{2}

n \sum r = 0 r^{2} . (^{n} C_{r})^{2}

r = n \sum r = 1 \frac{r^{4} + r^{2} + 1}{r^{4} + r} = \frac{675}{26},

n

n

\sum_{r = 0}^{n} \frac{{(- 1)}^{r}}{{^{n} C}_{r}}

x \in R, x \neq - 1

(1 + x)^{2016} + x (1 + x)^{2015} + x^{2} (1 + x)^{2014} + . . . + x^{2016} = 2016 \sum i = 0 a_{i} x^{i}

a_{17}

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