### Related Formula Many finite fractional sum questions involving functional terms rely on identifying an underlying symmetric summation invariant, typically of the form f(x) + f(k-x) = textconstant$f(x) + f(k-x) = \text{constant}$. ### Core Logic First simplify the expression for f(x)$f(x)$ algebraically: f(x) = frac4 cdot 2^x + 162 cdot (2^x)^2 + 16 cdot 2^x + 32$f(x) = \frac{4 \cdot 2^x + 16}{2 \cdot (2^x)^2 + 16 \cdot 2^x + 32}$ Factor out 4$4$ from the numerator and 2$2$ from the denominator: f(x) = frac4(2^x + 4)2[(2^x)^2 + 8 cdot 2^x + 16] = frac2(2^x + 4)(2^x + 4)^2 = frac22^x + 4$f(x) = \frac{4(2^x + 4)}{2[(2^x)^2 + 8 \cdot 2^x + 16]} = \frac{2(2^x + 4)}{(2^x + 4)^2} = \frac{2}{2^x + 4}$ ### Step 1: Establish Symmetry Pairings Let's check the value of f(x) + f(4-x)$f(x) + f(4-x)$: f(4-x) = frac22^4-x + 4 = frac2frac162^x + 4 = frac2 cdot 2^x16 + 4 cdot 2^x = frac2 cdot 2^x4(2^x + 4) = frac2^x2(2^x + 4)$f(4-x) = \frac{2}{2^{4-x} + 4} = \frac{2}{\frac{16}{2^x} + 4} = \frac{2 \cdot 2^x}{16 + 4 \cdot 2^x} = \frac{2 \cdot 2^x}{4(2^x + 4)} = \frac{2^x}{2(2^x + 4)}$ Now compute the sum directly: f(x) + f(4-x) = frac22^x + 4 + frac2^x2(2^x + 4) = frac4 + 2^x2(2^x + 4) = frac12$f(x) + f(4-x) = \frac{2}{2^x + 4} + \frac{2^x}{2(2^x + 4)} = \frac{4 + 2^x}{2(2^x + 4)} = \frac{1}{2}$ Hence, whenever two input arguments sum up to 4$4$, the sum of their functional values is exactly frac12$\frac{1}{2}$. ### Step 2: Group the Finite Series Terms Consider the terms inside the requested sequence: frac115 + frac5915 = frac6015 = 4 implies fleft(frac115 ight) + fleft(frac5915 ight) = frac12$\frac{1}{15} + \frac{59}{15} = \frac{60}{15} = 4 \implies f\left(\frac{1}{15} ight) + f\left(\frac{59}{15} ight) = \frac{1}{2}$ frac215 + frac5815 = frac6015 = 4 implies fleft(frac215 ight) + fleft(frac5815 ight) = frac12$\frac{2}{15} + \frac{58}{15} = \frac{60}{15} = 4 \implies f\left(\frac{2}{15} ight) + f\left(\frac{58}{15} ight) = \frac{1}{2}$ This complementary pairing continues up to: fleft(frac2915 ight) + fleft(frac3115 ight) = frac12$f\left(\frac{29}{15} ight) + f\left(\frac{31}{15} ight) = \frac{1}{2}$ This yields exactly 29$29$ distinct pairs. The single middle term left unpaired corresponds to: textMiddle Term = fleft(frac3015 ight) = f(2) = frac22^2 + 4 = frac28 = frac14$\text{Middle Term} = f\left(\frac{30}{15} ight) = f(2) = \frac{2}{2^2 + 4} = \frac{2}{8} = \frac{1}{4}$ ### Step 3: Evaluate Final Expression Compute the total value by multiplying the grouped sum by 8: textTotal = 8 cdot left[ 29 cdot left(frac12right) + frac14 right]$\text{Total} = 8 \cdot \left[ 29 \cdot \left(\frac{1}{2}\right) + \frac{1}{4} \right]$ textTotal = 8 cdot frac292 + 8 cdot frac14 = 116 + 2 = 118$\text{Total} = 8 \cdot \frac{29}{2} + 8 \cdot \frac{1}{4} = 116 + 2 = 118$ ### Pattern Recognition Whenever a symmetric set of arguments is presented inside a summation matching frackn + fracN-kn = textconstant$\frac{k}{n} + \frac{N-k}{n} = \text{constant}$, look for an algebraic reduction of f(x)$f(x)$ that yields a uniform constant sum for symmetric pairs. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Functions