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Let $a,b,c,\epsilon \: R$ if $f\left ( x \right )= ax^{2}+bx+c$ is such that a+b+c=3 and

$f(x+y)= f(x)+f(y)+xy$ $\forall \: x,y\: \epsilon \: R\: then$

$\sum_{n=1}^{10}\: f(n)$ is equal to :

Option 1)
165

Option 2)
190

Option 3)
255

Option 4)
330

Answers (2)

best_answer

As we learnt in

Summation of series of natural numbers -

$\sum_{k=1}^{n}K= \frac{1}{2}n\left ( n+1 \right )$

- wherein

Sum of first n natural numbers

$1+2+3+4+------+n= \frac{n(n+1)}{2}$

Summation of series of natural numbers -

$\sum_{k=1}^{n}K^{2}= \frac{1}{6}n\left ( n+1 \right )\left ( 2n+1 \right )$

- wherein

Sum of squares of first n natural numbers

$1^{2}+2^{2}+3^{2}+4^{2}+------+n^{2}= \frac{n(n+1)\left ( 2n+1 \right )}{6}$

f(x) = ax²+ bx + c

a + b + c = 3

f(x+y) = f(x) + f(y)+xy

Then $\sum_{n-1}^{10}f(n)$

f(1) = a + b + c =3

$\therefore$ f(1) =3

put y = 1

f(2)= f(1) + f(1)+1 $\times$ 1

=2f(1)+1

=2 $\times$ 3+1=7

put x=1, y = 2

f(3)= f(1)+f(2)+2 $\times$ 1

=3+7+2

=12

$\therefore$ f(1)+f(2)+(f3)+ ------= 3+7+12+ ---

Now

$T_{n}=$ An²+Bn+C

put n =1

3 = A+B+C

At n =2

7 = 4A+2B+C

At n =3

12 = 9A+3B+C

$\therefore$ 4 = 3A+B

5 = 5A+B

1= 2A

$\therefore A=\frac{1}{2}$

$\therefore$ B+C = $3-\frac{1}{2}=\frac{5}{2}$

$\therefore$ 2B+C

$= 7- 4\times\frac{1}{2}$

=7-2 = 5

$B=\frac{5}{2}$

C=0

So $T_{n}=\frac{n^{2}}{2}+\frac{5n}{2}$

$S_{n}=\sum T_{n}=\frac{1}{2}\cdot \frac{n\left ( n+1 \right )\left ( 2n+1 \right )}{6}+\frac{5}{2}\times\frac{n\left ( n+1 \right )}{2}$

$=\frac{n\left ( n+1 \right )\left ( 2n+1 \right )}{12}+\frac{5n\left ( n+1 \right )}{4}$

put n =10

$S_{10}= \frac{10\times11\times 21}{12}+\frac{50\times11}{4}$

$=\frac{55\times7}{2}+\frac{25\times11}{2}$

$=\frac{385+275}{2}$

$=\frac{660}{2}$

= 330

Option 1)

165

Incorrect option

Option 2)

190

Incorrect option

Option 3)

255

Incorrect option

Option 4)

330

Correct option

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Plabita

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