Let for some be a function satisfying for all . If f(0) = –1 and f(1) = 2, then the value of is: [2026]

Question

Let for some $α \in R, R \to R$ be a function satisfying $f (x + y) = f (x) + 2 y^{2} + y + α x y$ for all $x, y \in ℝ$ .

If f(0) = –1 and f(1) = 2, then the value of $\sum_{n = 1}^{5} (α + f (n))$ is: [2026]

Smart Achievers · Accepted Answer

(2)

$f (x + y) = f (x) + 2 y^{2} + y + α x y$ ; $x, y \in R$

Put x = 0, $f (y) = f (0) + 2 y^{2} + y + α (0) (y)$

$\Rightarrow f (y) = - 1 + 2 y^{2} + y$ [ $∵ f (0) = - 1$ ]

$\Rightarrow f (y) = 2 y^{2} + 1$ $∴ f (n) = 2 n^{2} + n - 1$

$\Rightarrow f (2) = 8 + 2 - 1 = 9$

Now, f(1) = 2

Put x = 1, y = 1; f(2) = f(1) + 2 + 1 + $α$

$\Rightarrow 9 = 5 + α$

$\Rightarrow α = 4$

Now, $\sum_{n = 1}^{5} (α + f (n)) = \sum_{n = 1}^{5} (4 + (2 n^{2} + n - 1))$

$= \sum_{n = 1}^{5} (2 n^{2} + n + 3)$

$= 2 \sum_{n = 1}^{5} n^{2} + \sum_{n = 1}^{5} n + 3 \sum_{n = 1}^{5}$

$= 2 \times [\frac{5 (6) \times 11}{6}] + \frac{5 \times 6}{2} + 3 \times 5$

$= 2 \times 55 + 15 + 15 = 140$

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