3. nC0−8. nC1+13. nC1−18. nC3+⋯+(n+1)terms=

3. nC0−8. nC1+13. nC1−18. nC3+⋯+(n+1)termstn+1=[3+(n).5] nCnlet k=3. nC0−8 nC1+13. nC2+⋯(−1)n(5n+3) nCnk=(−1)n(5n+3) nCn+(−1)n(5n−2) nCn−1+⋯+3. nC02k=(−1)n[(5n+6)[C0−C1+C2−C3+⋯]]2k=(−1)n(5n+6)(0) k=0

3. nC0−8. nC1+13. nC1−18. nC3+⋯+(n+1)terms=

3. nC0−8. nC1+13. nC1−18. nC3+⋯+(n+1)termstn+1=[3+(n).5] nCnlet k=3. nC0−8 nC1+13. nC2+⋯(−1)n(5n+3) nCnk=(−1)n(5n+3) nCn+(−1)n(5n−2) nCn−1+⋯+3. nC02k=(−1)n[(5n+6)[C0−C1+C2−C3+⋯]]2k=(−1)n(5n+6)(0) k=0

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$3.^{n} C_{0} - 8.^{n} C_{1} + 13.^{n} C_{1} - 18.^{n} C_{3} + \dots + (n + 1) terms =$

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Detailed Solution

$\begin{array}{l} 3.^{n} C_{0} - 8.^{n} C_{1} + 13.^{n} C_{1} - 18.^{n} C_{3} + \dots + (n + 1) terms \\ t_{n + 1} = [3 + (n) .5]^{n} C_{n} \\ let k = 3.^{n} C_{0} - 8^{n} C_{1} + 13.^{n} C_{2} + \dots {(- 1)}^{n} (5 n + 3)^{n} C_{n} \\ k = {(- 1)}^{n} (5 n + 3)^{n} C_{n} + {(- 1)}^{n} (5 n - 2)^{n} C_{n - 1} + \dots + 3.^{n} C_{0} \\ 2 k = {(- 1)}^{n} [(5 n + 6) [C_{0} - C_{1} + C_{2} - C_{3} + \dots]] \\ 2 k = {(- 1)}^{n} (5 n + 6) (0) \\ k = 0 \end{array}$