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)

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) = ax+ 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 \times1

=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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