If the last term in the binomial expansion of 31 - 5-1-3n is 1-56 - 5log 5 243- then the number of terms in the expansion is

T_r+1= ^nC_r a^n r b^r ∴ Last term of nbsp;(3^1/5 1√(3))^n is nbsp; T_n+1= ^nC_n(3^1/5)^n n( 1√(3))^n =( 1)^n3^n/3 Also, (15^6/5)^log_5243=1(5^6/5)^5log_53= ...

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Standard XII

Mathematics

Binomial Theorem

If the last t...

Question

If the last term in the binomial expansion of ${(3^{1 / 5} - \frac{1}{\sqrt[3]{3}})}^{n}$ is ${(\frac{1}{5^{6 / 5}})}^{{log}_{5} 243}$ , then the number of terms in the expansion is

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Solution

$T_{r + 1} =^{n} C_{r} a^{n - r} b^{r}$
$∴$ Last term of ${(3^{1 / 5} - \frac{1}{\sqrt[3]{3}})}^{n}$ is
$T_{n + 1} =^{n} C_{n} (3^{1 / 5})^{n - n} {(- \frac{1}{\sqrt[3]{3}})}^{n}$
$= \frac{(- 1)^{n}}{3^{n / 3}}$
Also,
${(\frac{1}{5^{6 / 5}})}^{{log}_{5} 243} = \frac{1}{(5^{6 / 5})^{5 {log}_{5} 3}} = 3^{- 6}$
Now,
$\frac{(- 1)^{n}}{3^{n / 3}} = 3^{- 6}$
$\Rightarrow n = 18$
Number of terms $= 19$

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If the last term in the binomial expansion of (31/5−13√3)n is (156/5)log5243, then the number of terms in the expansion is

Tr+1= nCr an−r br ∴ Last term of (31/5−13√3)n is Tn+1= nCn(31/5)n−n(−13√3)n =(−1)n3n/3 Also, (156/5)log5243=1(56/5)5log53=3−6 Now, (−1)n3n/3=3−6 ⇒n=18 Number of terms =19

If the last term in the binomial expansion of ${(3^{1 / 5} - \frac{1}{\sqrt[3]{3}})}^{n}$ is ${(\frac{1}{5^{6 / 5}})}^{{log}_{5} 243}$ , then the number of terms in the expansion is