Karnataka PUC I Binomial Theorem — practice questions — page 2

1. syllabus states the Binomial Theorem: $(a+b)^n = \sum_{r=0}^{n} \binom{n}{r}\,a^{n-r}\,b^r$. 2. Note: option B's formula $\sum \binom{n}{r}\,a^r\,b^{n-r}$ is also correct because $\binom{n}{r} = \binom{n}{n-r}$ (symmetric). However, A is the standard syllabus form. 3. Coefficients $\binom{n}{r}$ are called BINOMIAL COEFFICIENTS, and they appear in PASCAL'S triangle row $n$. 4. The expansion has $n+1$ terms (one for each value $r = 0, 1, \ldots, n$). 5. Options C, D are missing most of the terms — only first and last, or a wrong partial form. _Source: syllabus Mathematics, Ch 7 "Binomial Theorem", §7.2 (Statement of the theorem), p. 1–3._

1. The sum of binomial coefficients is $\binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \ldots + \binom{n}{n}$. 2. To get this sum, substitute $x = 1$ in the expansion $(1+x)^n = \sum_{r=0}^{n}\binom{n}{r}\,x^r$. 3. LHS becomes $(1+1)^n = 2^n$. RHS becomes $\sum \binom{n}{r}$. 4. So $\sum_{r=0}^{n}\binom{n}{r} = 2^n$. 5. Example check: for $n = 3$, $\binom{3}{0} + \binom{3}{1} + \binom{3}{2} + \binom{3}{3} = 1 + 3 + 3 + 1 = 8 = 2^3$. ✓ 6. Option A ($n^2$) is wrong (only matches for $n = 2, 4$). Option C is half. Option D is the number of TERMS, not the sum. _Source: syllabus Mathematics, Ch 7, §7.2.2 (Some special cases — $x = 1$), p. 5._

1. Your syllabus: writing out the binomial expansion term by term ($r = 0, 1, 2, \ldots$): - $T_1$: $r = 0$, term = $\binom{n}{0}\,a^n\,b^0$ - $T_2$: $r = 1$, term = $\binom{n}{1}\,a^{n-1}\,b^1$ - $T_{r+1}$: $r$-th index, term = $\binom{n}{r}\,a^{n-r}\,b^r$. 2. So the GENERAL term is $T_{r+1} = \binom{n}{r}\,a^{n-r}\,b^r$. 3. Important: the index of $T$ is $r+1$ (not $r$) — the very first term corresponds to $r=0$. 4. Option B has the powers swapped (wrong unless $b$ is the leading term). Option C has wrong binomial coefficient. Option D omits the binomial coefficient. _Source: syllabus Mathematics, Ch 7, §7.2.1 (General Term, Eq. for $T_{r+1}$), p. 4._