GO Classes CS/DA 2025 | Weekly Quiz 1 | Fundamental Course | Question: 1

Answer 1

If you simply visualize this question in the following way, it becomes very easy.

S = 1+2+3+...+n

A = 1+2+3+...+a

B = a+...+n

Now option (a), A + B = (1+2+3+...+a) + (a+...+n) ,so you see a gets counted twice thus this option is wrong.

Now option (b), = (1+2+3+...+a-1) + a + (a+...+n) ,here also a gets counted twice so this one is wrong too.

Now option (c), =  (1+2+3+...+a-1) + (a+...+n) ,this one is a continuous series hence it's correct.

Now option (d), A + B – f(a) = (1+2+3+...+a) + (a+...+n) – a ,this is correct as the extra a is subtracted.

 

Hence correct options are (c) and (d)

Comments

$i$ needs not come from a set of consecutive integers. You have written a delimited form for the sigma notation.

$\displaystyle \Sigma$ is considered as a discrete operator and so, index variable can take values from a discrete set.

Formally, when we write \[ \sum_{P(k)}^{} a_k  \]

It means we are doing sum over all terms $a_k$ such that $k$ is an integer which satisfying a given property $P(k)$ where $P(k)$ is any statement about $k$ which can either be true or false.

So, suppose here, for $A,$ $P(k)$ is defined as: index variable $i$ comes from a set of primes which are less than or equal to $7$ i.e. $\pi(7).$ So, $A= f(x_2)+f(x_3)+f(x_5)+f(x_7)$ and

suppose for $B,$ $P(k)$ is defined as: index variable $i$ comes from a set of positive even integers which are greater than or equal to $7.$ So, $A= f(x_8)+f(x_{10})+f(x_{12})+...+f(x_n)$ where $n$ is an even integer.

Now, does upper limit of $A$ is same as lower limit of $B$ ?

And now, which option would be correct now ?

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@chokostar 
You have assumed f(x) = x (like f(x=1) = 1, f(x=2) = 2 and so on) and this is wrong.

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Hello @Vineeth Rambhiya, I have not considered so. I have simply used the sigma notation to expand S, A and B.

Answer 2

Thanks you!!

Answer 3

${S}$=$\sum_{i=1}^{n}$  ${f(x}$$_{i}$${)}$ = ${f(x}$$_{1}$${)}$${+}$${f(x}$$_{2}$${)}$${+}$${… +f(x}$$_{a}$${)+}$${….+}$${f(x}$$_{n}$${)}$

${A}$=$\sum_{i=1}^{a}$  ${f(x}$$_{i}$${)}$ = ${f(x}$$_{1}$${)}$${+}$${f(x}$$_{2}$${)}$${+}$${… +f(x}$$_{a}$${)}$

${B}$=$\sum_{i=a}^{n}$  ${f(x}$$_{i}$${)}$ = ${f(x}$$_{a}$${)}$${+}$${f(x}$$_{a+1}$${)}$${+}$${… +f(x}$$_{n}$${)}$

now let see options

• A.  ${S=A+B}$
  
   ${A+B=}$  ${f(x}$$_{1}$${)}$${+}$${f(x}$$_{2}$${)}$${+}$${… +f(x}$$_{a}$${)+}$ ${f(x}$$_{a}$${)}$${+}$${f(x}$$_{a+1}$${)}$${+}$${… +f(x}$$_{n}$${)}$ 
                 ${=f(x}$$_{1}$${)}$${+}$${f(x}$$_{2}$${)}$${+}$${… +f(x}$$_{a}$${)+}$${….+}$${f(x}$$_{n}$${)}$${+f(x}$$_{a}$${)}$
                 ${=f(x}$$_{1}$${)}$${+S}$
  
  hence false
   
• B.   ${S=}$$\sum_{i=1}^{a-1}$  ${f(x}$$_{i}$${)}$${+}$${f(x}$$_{a}$${)}$${+B}$
  
  ${R.H.S =}$$\sum_{i=1}^{a-1}$  ${f(x}$$_{i}$${)}$${+}$${f(x}$$_{a}$${)}$${+B}$
                  ${=f(x}$$_{1}$${)}$${+}$${f(x}$$_{2}$${)}$${+}$${… +f(x}$$_{a-1}$${)+}$${f(x}$$_{a}$${)+}$${B}$
                  ${=A+B}$
  
  Since ${S}$$\neq$${A+B}$
  
  Hence False.
• C. ${S=}$$\sum_{i=1}^{a-1}$  ${f(x}$$_{i}$${)}$${+}$${B}$
  
   ${R.H.S =}$$\sum_{i=1}^{a-1}$  ${f(x}$$_{i}$${)}$${+}$${B}$
                   ${=f(x}$$_{1}$${)}$${+}$${f(x}$$_{2}$${)}$${+}$${… +f(x}$$_{a-1}$${)+}$${f(x}$$_{a}$${)}$${+}$${f(x}$$_{a+1}$${)}$${+}$${… +f(x}$$_{n}$${)=S}$
  
  Hence  ${S=}$$\sum_{i=1}^{a-1}$  ${f(x}$$_{i}$${)}$${+}$${B}$  True
   
• D.  ${S=A+B}$${-f(x}$$_{a}$${)}$
  
    ${R.H.S =}$${A+B}$${-f(x}$$_{a}$${)}$
                    ${=f(x}$$_{1}$${)}$${+}$${f(x}$$_{2}$${)}$${+}$${… +f(x}$$_{a}$${)+}$ ${f(x}$$_{a}$${)}$${+}$${f(x}$$_{a+1}$${)}$${+}$${… +f(x}$$_{n}$${)}$${-f(x}$$_{a}$${)}$
                   ${=f(x}$$_{1}$${)}$${+}$${f(x}$$_{2}$${)}$${+}$${… +f(x}$$_{a}$${)+}$${….+}$${f(x}$$_{n}$${)}$${+f(x}$$_{a}$${)}$${-f(x}$$_{a}$${)}$
                  ${=f(x}$$_{1}$${)}$${+}$${f(x}$$_{2}$${)}$${+}$${… +f(x}$$_{a}$${)+}$${….+}$${f(x}$$_{n}$${)=S}$
  
  Hence ${S=A+B}$${-f(x}$$_{a}$${)}$ True

${Hence}$  ${ans : C,D}$

Comments

c,d are correct

Answer 4

c & d